Binomial Theorem
Explanation:
$ \begin{aligned} & \frac{1}{{ }^n C_r}+\frac{1}{{ }^n C_{r+1}}=\frac{{ }^n C_{r+1}+{ }^n C_r}{{ }^n C_r \cdot{ }^n C_{r+1}}=\frac{{ }^{n+1} C_{r+1}}{{ }^n C_r \cdot{ }^n C_{r+1}}=\frac{\frac{n+1}{r+1}+{ }^n C_r}{{ }^n C_r \cdot{ }^n C_{r+1}}=\frac{(n+1)}{(r+1) \cdot \frac{n}{r+1} \cdot{ }^{n-1} C_r}=\frac{n+1}{n \cdot{ }^{n+1} C_r} \\ & \therefore\left(\frac{1}{15 C_0}+\frac{1}{15 C_1}\right)\left(\frac{1}{15 C_1}+\frac{1}{15 C_2}\right) \cdots\left(\frac{1}{15 C_{12}}+\frac{1}{15 C_{13}}\right)=\frac{16}{15 \cdot{ }^{14} C_0} \cdot \frac{16}{15 \cdot{ }^{14} C_1} \cdot \frac{16}{15 \cdot{ }^{14} C_{12}}=\frac{\left(\frac{16}{15}\right)^{13}}{{ }^{14} C_0 \cdot{ }^{14} C_1 \cdot{ }^{14} C_2 \cdot \cdots{ }^{14} C_{12}} \\ & \therefore \alpha=\frac{16}{15} \\ & \therefore 30 \alpha=32 \end{aligned} $
If $\left(1-x^3\right)^{10}=\sum\limits_{\mathrm{r}=0}^{10} \mathrm{a}_{\mathrm{r}} x^{\mathrm{r}}(1-x)^{30-2 \mathrm{r}}$, then $\frac{9 \mathrm{a}_9}{\mathrm{a}_{10}}$ is equal to $\_\_\_\_$ .
Explanation:
Given
$ (1-x^3)^{10}=\sum_{r=0}^{10} a_r x^r(1-x)^{30-2r}. $
We have to find
$ \frac{9a_9}{a_{10}}. $
Let us simplify the left side first.
Since
$ 1-x^3=(1-x)(1+x+x^2), $
so
$ (1-x^3)^{10}=(1-x)^{10}(1+x+x^2)^{10}. $
Also,
$ 1+x+x^2=\frac{1-x^3}{1-x}. $
Now notice that
$ 1+x+x^2=(1-x)^ {-2}\big(x+(1-x)^2\big)? $
This is not the most useful form here. So let us proceed in a better NCERT-style way.
We are given
$ (1-x^3)^{10}=\sum_{r=0}^{10} a_r x^r(1-x)^{30-2r}. $
Take out $(1-x)^{10}$ from the left side:
$ (1-x^3)^{10}=(1-x)^{10}(1+x+x^2)^{10}. $
Hence
$ (1+x+x^2)^{10} =\sum_{r=0}^{10} a_r x^r(1-x)^{20-2r}. $
Now observe that
$ 1+x+x^2=(1-x)^2+3x. $
Because
$ (1-x)^2+3x=1-2x+x^2+3x=1+x+x^2. $
Therefore
$ (1+x+x^2)^{10}=\big((1-x)^2+3x\big)^{10}. $
Using binomial theorem,
$ \big((1-x)^2+3x\big)^{10} =\sum_{r=0}^{10} \binom{10}{r}(3x)^r\big((1-x)^2\big)^{10-r}. $
So
$ (1+x+x^2)^{10} =\sum_{r=0}^{10} \binom{10}{r}3^r x^r(1-x)^{20-2r}. $
Comparing this with
$ (1+x+x^2)^{10} =\sum_{r=0}^{10} a_r x^r(1-x)^{20-2r}, $
we get
$ a_r=\binom{10}{r}3^r. $
Hence
$ a_9=\binom{10}{9}3^9=10\cdot 3^9, $
and
$ a_{10}=\binom{10}{10}3^{10}=3^{10}. $
Therefore
$ \frac{9a_9}{a_{10}} =9\cdot \frac{10\cdot 3^9}{3^{10}} =9\cdot \frac{10}{3} =30. $
So the required value is
$ \boxed{30}. $
If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^3}-x^4\right)^n, x \neq 0$, is zero, then the value of $n$ is $\_\_\_\_$ .
Explanation:
For
$ \left(\frac{1}{x^3}-x^4\right)^n = \left(x^{-3}-x^4\right)^n, $
let us write the general term in the expansion.
The general term is
$ T_{r+1}=\binom{n}{r}(x^{-3})^{\,n-r}(-x^4)^r $
$ =\binom{n}{r}(-1)^r x^{-3(n-r)}x^{4r} $
$ =\binom{n}{r}(-1)^r x^{-3n+3r+4r} $
$ =\binom{n}{r}(-1)^r x^{-3n+7r}. $
So, the coefficient of $x^k$ comes from the term for which
$ -3n+7r=k. $
Now find the coefficients of $x^7$ and $x^{14}$.
For $x^7$:
$ -3n+7r=7 $
$ 7r=3n+7 $
$ r=\frac{3n+7}{7}. $
For $x^{14}$:
$ -3n+7r=14 $
$ 7r=3n+14 $
$ r=\frac{3n+14}{7}. $
Since $r$ must be an integer, $3n+7$ and $3n+14$ must both be divisible by $7$.
Now,
$ 3n+7\equiv 0 \pmod{7} \Rightarrow 3n\equiv 0 \pmod{7}. $
Since $3$ and $7$ are coprime,
$ n\equiv 0 \pmod{7}. $
So let
$ n=7m. $
Then the two corresponding values of $r$ are
For $x^7$:
$ r_1=\frac{3(7m)+7}{7}=3m+1. $
For $x^{14}$:
$ r_2=\frac{3(7m)+14}{7}=3m+2. $
Hence the coefficients are
$ \binom{n}{3m+1}(-1)^{3m+1} \quad \text{and} \quad \binom{n}{3m+2}(-1)^{3m+2}. $
Their sum is zero, so
$ \binom{n}{3m+1}(-1)^{3m+1}+\binom{n}{3m+2}(-1)^{3m+2}=0. $
Taking common sign,
$ -\binom{n}{3m+1}+\binom{n}{3m+2}=0 $
or
$ \binom{n}{3m+2}=\binom{n}{3m+1}. $
Using the NCERT result:
$ \binom{n}{r}=\binom{n}{r+1} \Rightarrow r+1=\frac{n}{2} $
or equivalently,
$ 2r+1=n. $
Here,
$ r=3m+1. $
So,
$ 2(3m+1)+1=7m $
$ 6m+2+1=7m $
$ 6m+3=7m $
$ m=3. $
Therefore,
$ n=7m=21. $
Hence, the value of $n$ is
$ \boxed{21}. $
Explanation:
$\begin{aligned} & (1919)^{1919}=(1920-1)^{1919} \\ & ={ }^{1919} \mathrm{C}_0(1920)^{1919}-{ }^{1919} \mathrm{C}_1(1920)^{1918}+\ldots . \\ & +{ }^{1919} \mathrm{C}_{1918}(1920)^1-{ }^{1919} \mathrm{C}_{1919} \\ & =100 \lambda+1919 \times 1920-1 \\ & =100 \lambda+3684480-1 \\ & =100 \lambda+\ldots \ldots \ldots . .79 \text { (last two digit) } \\ & \Rightarrow \text { Number having last two digit } 79 \\ & \therefore \text { Product of last two digit } 63 \end{aligned}$
Explanation:
$\begin{aligned} &\begin{aligned} & (1-\mathrm{x})^{20}={ }^{20} \mathrm{C}_0-{ }^{20} \mathrm{C}_1 \mathrm{x}+{ }^{20} \mathrm{C}_2 \mathrm{x}^2 \ldots . .+{ }^{20} \mathrm{C}_{20} \mathrm{x}^{20} \\ & \frac{(1-\mathrm{x})^{20}}{\mathrm{x}^2}=\frac{{ }^{20} \mathrm{C}_0}{\mathrm{x}^2}-\frac{{ }^{20} \mathrm{C}_1}{\mathrm{x}}+{ }^{20} \mathrm{C}_2-{ }^{20} \mathrm{C}_3 \mathrm{x}+{ }^{20} \mathrm{C}_4 \mathrm{x}^2 \ldots . \end{aligned}\\ &\text { Diff twice and put } \mathrm{x}=1\\ &\begin{aligned} & =6-{ }^{20} \mathrm{C}_1(2)+\mathrm{A} \\ & \mathrm{~A}=40-6=34 \end{aligned} \end{aligned}$
Let $\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$. If $\left(a_1+a_3+a_5+\ldots+a_{19}\right)-11 a_2=121 k$, then $k$ is equal to_________ .
Explanation:
Let $f(x)=\left(1+x+x^2\right)^{10}=\sum_{r=0}^{20} a_r x^r$
The sum of odd coefficients: $S_{\text {odd }}=a_1+a_3+a_5+\cdots$ $+a_{19}$
Subtracting $11 a_2$ from above will give the answer
$\begin{aligned} & S_{\text {odd }}=\frac{f(1)-f(-1)}{2} \\ & f(1)=(1+1+1)^{10}=3^{10} \\ & f(-1)=(1-1+1)^{10}=(1)^{10}=1 \\ & S_{\text {odd }}=\sum_{\text {odd } r} a_r=\frac{3^{10}-1}{2} \end{aligned}$
Now for $a_2$
$1+x+x^2=\frac{1-x^3}{1-x} \Rightarrow f(x)=\left(\frac{1-x^3}{1-x}\right)=\frac{\left(1-x^3\right)^{10}}{(1-x)^{10}}$
Now use:
$\begin{aligned} & \left(1-x^3\right)^{10}=\sum_{x=0}^{10}(-1)^k\binom{10}{k} x^{3 k} \\ & (1-x)^{-10}=\sum_{r=0}^{\infty}\binom{r+9}{9} x^r \end{aligned}$
So
$f(x)=\left(\sum_{k=0}^{10}(-1)^k\binom{10}{k} x^{3 k}\right) \cdot\left(\sum_{r=0}^{\infty}\binom{r+9}{9} x^r\right)$
Only the term with $x^0$ from the first sum (i.e., $k=0$ ) can contribute to $x^2$, since all other $k \geq 1$ gives $x^{3 k} \geq$ $x^3$
From $\left(1-x^3\right)^{10}$ : the $x^0$ term is $\binom{10}{0}=1$
From $(1-x)^{-10}$ : the coefficient of $x^2$ is
$\binom{2+9}{9}=\binom{11}{9}=55$
Hence, $a_2=1.55=55$
$\begin{aligned} & \text { Now, } S_{\text {odd }}-11 a_2=\frac{3^{10}-1}{2}-11 \cdot 55=121 k \\ & 3^{10}=59049 \end{aligned}$
So:
$\begin{aligned} & S=\frac{59049-1}{2}-605=\frac{59048}{2}-605 \\ & =29524-605=28919 \end{aligned}$
So:
$121 k=28919 \Rightarrow k=\frac{28919}{121}=239$
If $\alpha=1+\sum\limits_{r=1}^6(-3)^{r-1} \quad{ }^{12} \mathrm{C}_{2 r-1}$, then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x-\sqrt{3} y+1=0$ is ________.
Explanation:
$\begin{aligned} &\begin{aligned} \alpha & =1+\sum_{\mathrm{r}=1}^6(-1)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} 3^{\mathrm{r}-1} \\ \alpha & =1+\sum_{\mathrm{r}=1}^6{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} \frac{(\sqrt{3} \mathrm{i})^{2 \mathrm{t}-1}}{\sqrt{3} \mathrm{i}} \quad \mathrm{i}=\text { iota, let } \sqrt{3} \mathrm{i}=\mathrm{x} \\ \alpha & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left({ }^{12} \mathrm{C}_1 \mathrm{x}+{ }^{12} \mathrm{C}_3 \mathrm{x}^3+\ldots .{ }^{12} \mathrm{C}_{11} \mathrm{x}^{11}\right) \\ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{(1+\sqrt{3} \mathrm{i})^{12}-(1-\sqrt{3} \mathrm{i})^{12}}{2}\right) \\ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{\left(-2 \omega^2\right)^{12}-(2 \omega)^{12}}{2}\right)=1 \end{aligned}\\ &\text { so distance of }(12, \sqrt{3}) \text { from } x-\sqrt{3} y+1=0 \text { is }\\ &\frac{12-3+1}{2}=5 \end{aligned}$
The sum of all rational terms in the expansion of $\left(1+2^{1 / 3}+3^{1 / 2}\right)^6$ is equal to _________.
Explanation:
$\left(1+2^{\frac{1}{3}}+3^{\frac{1}{2}}\right)^6$
$ = {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, {{r_1}} \,}} \right. \left| \!{\underline {\, {{r_2}} \,}} \right. \left| \!{\underline {\, {{r_3}} \,}} \right. }}{(1)^{{r_1}}}{(2)^{{{{r_2}} \over 3}}}{(3)^{{{{r_3}} \over 2}}}$
$\begin{array}{l|l|l} \mathrm{r}_1 & \mathrm{r}_2 & \mathrm{r}_3 \\ \hline 6 & 0 & 0 \\ 4 & 0 & 2 \\ 2 & 0 & 4 \\ 0 & 0 & 6 \\ \hline 3 & 3 & 0 \\ 1 & 3 & 2 \\ \hline 0 & 6 & 0 \\ \hline \end{array}$
$ = {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 6 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. }} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 4 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 2 \,}} \right. }}(3) + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 2 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 4 \,}} \right. }}{(3)^2} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 6 \,}} \right. }}{(3)^3} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 3 \,}} \right. \left| \!{\underline {\, 3 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. }}(2) + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 1 \,}} \right. \left| \!{\underline {\, 3 \,}} \right. \left| \!{\underline {\, 2 \,}} \right. }}{(2)^1}{(3)^1} + {{\left| \!{\underline {\, 6 \,}} \right. } \over {\left| \!{\underline {\, 0 \,}} \right. \left| \!{\underline {\, 6 \,}} \right. \left| \!{\underline {\, 0 \,}} \right. }}{(2)^2}$
$=1+45+135+27+40+360+4=612$
If $\sum_\limits{r=1}^{30} \frac{r^2\left({ }^{30} C_r\right)^2}{{ }^{30} C_{r-1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to _________.
Explanation:
$\begin{aligned} & \sum_{\mathrm{r}=1}^{30} \frac{\mathrm{r}^2\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)^2}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}} \\ & =\sum_{\mathrm{r}=1}^{30} \mathrm{r}^2\left(\frac{31-\mathrm{r}}{\mathrm{r}}\right) \cdot \frac{30!}{\mathrm{r}!(30-\mathrm{r})!} \\ & \left(\because \frac{{ }^{30} \mathrm{C}_{\mathrm{r}}}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}=\frac{30-\mathrm{r}+1}{\mathrm{r}}=\frac{31-\mathrm{r}}{\mathrm{r}}\right) \\ & =\sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 30!}{(\mathrm{r}-1)!(30-\mathrm{r})!} \\ & =30 \sum_{\mathrm{r}=1}^{30} \frac{(31-\mathrm{r}) 29!}{(\mathrm{r}-1)!(30-\mathrm{r})!} \\ & =30 \sum_{\mathrm{r}=1}^{30}(30-\mathrm{r}+1)^{29} \mathrm{C}_{30-\mathrm{r}} \\ & =30\left(\sum_{\mathrm{r}=1}^{30}(31-\mathrm{r})^{29} \mathrm{C}_{30-\mathrm{r}}+\sum_{\mathrm{r}=1}^{30}{ }^{29} \mathrm{C}_{30-\mathrm{r}}\right) \\ & =30\left(29 \times 2^{28}+2^{29}\right)=30(29+2) 2^{28} \\ & =15 \times 31 \times 2^{29} \\ & =465\left(2^{29}\right) \\ & \alpha=465 \end{aligned}$
If $\sum_\limits{r=0}^5 \frac{{ }^{11} C_{2 r+1}}{2 r+2}=\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}-\mathrm{n}$ is equal to __________.
Explanation:
$\begin{aligned} & (1+x)^{11}={ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11} \\ & \int_0^1(1+x)^{11} d x=\int_0^1\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x{ }^2+\cdots+{ }^{11} C_{11} x{ }^{11}\right) d x \\ & \left.\left.\frac{(1-x)^{12}}{12}\right]_0^1={ }^{11} C_{0 x}+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_0^1\end{aligned}$
$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{11}}{12} \cdots$ ........(1)
Now,
$ \begin{aligned} & \int_{-1}^0(1+x)^{11} d x=\int_{-1}^0\left({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^2+\cdots+{ }^{11} C_{11} x^{11}\right) d x \\ & \left.\left.\frac{(1+x)^{12}}{12}\right]_{-1}^0={ }^{11} C_0 x+\frac{{ }^{11} C_1 x^2}{2}+\frac{{ }^{11} C_2 x^3}{3}+\cdots+\frac{{ }^{11} C_{11} x^{12}}{12}\right]_{-1}^0 \end{aligned} $
$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3} \cdots$ …….(2)
$\begin{aligned} & \text { (1) }-(2) \\\\ & =\frac{2^{12}-2}{12}=2\left[\frac{C_1}{2}+\frac{C_3}{4}+\cdots\right] \\\\ & \Rightarrow \sum_{r=0}^5 \frac{C_{2 r+1}}{2 r+2}=\frac{2^{11}-1}{12}=\frac{2047}{12}=\frac{m}{n} \\\\ & =2047-12=2035\end{aligned}$
The remainder when $428^{2024}$ is divided by 21 is __________.
Explanation:
$\begin{aligned} & 428=21 \times 20+8 \\ \Rightarrow \quad & (428)^{2024} \equiv(20 \times 21+8)^{2024} \equiv 8^{2024}(\bmod 21) \\ & 8^2=21 \times 3+1 \\ & 8^{2024}=(21 \times 3+1)^{1012} \\ \Rightarrow \quad & 8^{2024} \equiv(21 \times 3+1)^{1012}(\bmod 21) \\ & \equiv 1^{2012}(\bmod 21) \\ & 428^{2024} \equiv 1(\bmod 21) \end{aligned}$
If the second, third and fourth terms in the expansion of $(x+y)^n$ are 135, 30 and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to __________.
Explanation:
$\begin{aligned} & T_2={ }^n C_1 y^1 \cdot x^{n-1}=135 \\ & T_3={ }^n C_2 y^2 \cdot x^{n-2}=30 \\ & T_4={ }^n C_3 y^3 x^{n-3}=\frac{10}{3} \end{aligned}$
$\begin{aligned} \Rightarrow \frac{135}{30} & =\left(\frac{x}{y}\right) \frac{n \cdot 2}{n(n-1)}=\left(\frac{2}{n-1}\right)\left(\frac{x}{y}\right) \quad \text{... (i)}\\ \frac{30}{\frac{10}{3}} & =\frac{n(n-1)}{2} \frac{3!}{n(n-1)(n-2)}\left(\frac{x}{y}\right) \\ 9 & =\left(\frac{3}{n-2}\right)\left(\frac{x}{y}\right) \end{aligned}$
$\begin{aligned} \Rightarrow & 3(n-2)=\frac{135}{60}(n-1) \Rightarrow n=5 \\ \Rightarrow & x=9 y \quad \text{.... (i)}\\ & y \cdot x^4=27 \Rightarrow \frac{x}{9} \cdot x^4=3^3 \\ \Rightarrow & x^5=3^5 \Rightarrow x=3 y=\frac{1}{3} \\ \Rightarrow & 6\left(5^3+3^2+\frac{1}{3}\right)=6\left(125+9+\frac{1}{3}\right) \end{aligned}$
$=6(134)+2=806$
If the constant term in the expansion of $\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$ is $\mathrm{p}$, then $108 \mathrm{p}$ is equal to ________.
Explanation:
$\text { General term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
$T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}$
Constant term in expansion of $\left(1+2 x-3 x^3\right)$
$\begin{aligned} & \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\ & =T_7-3 T_8={ }^9 C_6 3^{-3} \cdot 2^{-3}+3{ }^9 C_7 \cdot 3^{-5} \cdot 2^{-2} \\ & =\frac{3 \times 4 \times 7}{3^3 \cdot 2^3}+\frac{3 \times 9 \times 4}{3^5 \times 2^2}= \\ & p=\frac{42+12}{108}=\frac{54}{108} \\ & 108 p=54 \end{aligned}$
Let $a=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+...., \mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+....$ Then $\frac{2 b}{a^2}$ is equal to _________.
Explanation:
$\begin{aligned} & a=1+\frac{{ }^2 C_2}{3!}+\frac{{ }^3 C_2}{4!}+\frac{{ }^4 C_2}{5!}+\ldots \\ & b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{ }^2 C_2}{2!}+\ldots \\ & b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2}{3!}+\ldots=e^2 \end{aligned}$
Using $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x}{3!}+\ldots$
$\begin{aligned} a= & 1+\sum_{r=2}^{\infty} \frac{{ }^r C_2}{(r+1)!}=1+\sum_{r=2} \frac{r(r-1)}{2(r+1)!} \\ & =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{(r+1) r-2 r}{(r+1)!} \\ & =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-1)!}-\frac{1}{2} \sum_{r=2} \frac{2 r}{(r+1)!} \\ & =1+\frac{1}{2}\left(\frac{1}{1!}+\frac{1}{2!}+\ldots\right)-\sum_{r=2}^{\infty} \frac{(r+1)-1}{(r+1)!} \\ & =1+\frac{1}{2}(e-1)-\sum_{r=2}^{\infty} \frac{1}{r!}+\sum_{r=2} \frac{1}{(r+1)!} \\ & =1+\frac{1}{2}(e-1)-\left(e-\frac{1}{1!}-\frac{1}{0!}\right)+\left(e-\frac{1}{1!}-\frac{1}{0!}-\frac{1}{2!}\right) \\ & =1+\frac{e}{2}-\frac{1}{2}-e+2+e-2-\frac{1}{2}=\frac{e}{2} \\ \Rightarrow & \frac{2 b}{a^2}=\frac{2}{\frac{e^2}{4}} 8 \end{aligned}$
Explanation:
$ \text { Case-I : } \begin{array}{|c|c|c|} \hline \mathrm{r}_1 & \mathrm{r}_2 & \mathrm{r}_3 \\ \hline 0 & 6 & 8 \\ \hline 2 & 5 & 8 \\ \hline 4 & 4 & 8 \\ \hline 6 & 3 & 8 \\ \hline \end{array} $
$r_1+2 r_2=12 \quad\left(\right.$ Taking $\left.r_3=8\right)$
$ \begin{aligned} &\text { Case-II :}\\\\ &\begin{array}{|l|l|l|} \hline r_1 & r_2 & r_3 \\ \hline 1 & 7 & 7 \\ \hline 3 & 6 & 7 \\ \hline 5 & 5 & 7 \\ \hline \end{array} \end{aligned} $
$r_1+2 r_2=15\left(\right.$ Taking $\left.r_3=7\right)$
$ \begin{aligned} &\text { Case-III : }\\\\ &\begin{array}{|l|l|l|} \hline r_1 & r_2 & r_3 \\ \hline 4 & 7 & 6 \\ \hline 6 & 6 & 6 \\ \hline \end{array} \end{aligned} $
$\begin{aligned} & \text { Coefficient}=7+(15 \times 21)+(15 \times 35)+(35) \\\\ & -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\\\ & +(7 \times 28)=-678=\alpha \\\\ & |\alpha|=678\end{aligned}$
Let the coefficient of $x^r$ in the expansion of $(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}$ be $\alpha_r$. If $\sum_\limits{r=0}^n \alpha_r=\beta^n-\gamma^n, \beta, \gamma \in \mathbb{N}$, then the value of $\beta^2+\gamma^2$ equals _________.
Explanation:
$\begin{aligned} & (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\ & (x+2)^2+\ldots \ldots .+(x+2)^{n-1} \\ & \sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} \\ & =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots .+\left(\frac{3}{4}\right)^{n-1}\right] \\ & =4^{n-1} \times \frac{1-\left(\frac{3}{4}\right)^n}{1-\frac{3}{4}} \\ & =4^n-3^n=\beta^n-\gamma^n \\ & \beta=4, \gamma=3 \\ & \beta^2+\gamma^2=16+9=25 \end{aligned}$
In the expansion of $(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to __________.
Explanation:
$\begin{aligned} & (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\ & =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \end{aligned}$
$=\operatorname{coeff}\left(\mathrm{x}^3\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^{18}\right)$ in
$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =0-1 \\ & =-1 \end{aligned}$
$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$ in
$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =\left(\begin{array}{c} 17 \\ 2 \end{array}\right)-\left(\begin{array}{c} 17 \\ 1 \end{array}\right) \\ & =17 \times 8-17 \\ & =17 \times 7 \\ & =119 \end{aligned}$
Hence Answer $=119-1=118$
Let $\alpha=\sum_\limits{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_\limits{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$ If $5 \alpha=6 \beta$, then $n$ equals _______.
Explanation:
$\begin{aligned} \alpha= & \sum_{k=0}^n \frac{{ }^n C_k \cdot{ }^n C_k}{k+1} \cdot \frac{n+1}{n+1} \\ & =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} \\ \alpha & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} \\ \beta & =\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} \\ & \frac{1}{n+1} \sum_{k=0}^{n-1}{ }^n C_{n-k} \cdot{ }^{n+1} C_{k+2} \\ & =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} \\ \frac{\beta}{\alpha} & =\frac{2 n+1}{2 n+1} C_{n+2} \\ \frac{\beta}{\alpha} & =\frac{2 n+1-(n+2)+1}{n+2}=\frac{5}{6} \\ n & =10 \end{aligned}$
$\text { Number of integral terms in the expansion of }\left\{7^{\left(\frac{1}{2}\right)}+11^{\left(\frac{1}{6}\right)}\right\}^{824} \text { is equal to _________. }$
Explanation:
General term in expansion of $\left((7)^{1 / 2}+(11)^{1 / 6}\right)^{824}$ is $\mathrm{t}_{\mathrm{r}+1}={ }^{824} \mathrm{C}_{\mathrm{r}}(7)^{\frac{824-\mathrm{r}}{2}}(11)^{\mathrm{r} / 6}$
For integral term, $r$ must be multiple of 6.
Hence $r=0,6,12, ....... 822$
Remainder when $64^{32^{32}}$ is divided by 9 is equal to ________.
Explanation:
Let $32^{32}=\mathrm{t}$
$\begin{aligned} & 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 t} \\ & =9 \mathrm{k}+1 \end{aligned}$
Hence remainder $=1$
$\text { If } \frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots+\frac{{ }^{11} C_9}{10}=\frac{n}{m} \text { with } \operatorname{gcd}(n, m)=1 \text {, then } n+m \text { is equal to }$ _______.
Explanation:
$\begin{aligned} & \sum_{\mathrm{r}=1}^9 \frac{{ }^{11} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1} \\ & =\frac{1}{12} \sum_{\mathrm{r}=1}^9{ }^{12} \mathrm{C}_{\mathrm{r}+1} \\ & =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} \\ & \therefore \mathrm{m}+\mathrm{n}=2041 \end{aligned}$
The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}\left(1+x+x^2\right)^{2007}$ is equal to _________.
Explanation:
$\begin{aligned} & (1-x)(1-x)^{2007}\left(1+x+x^2\right)^{2007} \\ & (1-x)\left(1-x^3\right)^{2007} \\ & (1-x)\left({ }^{2007} C_0-{ }^{2007} C_1\left(x^3\right)+\ldots \ldots .\right) \end{aligned}$
General term
$\begin{aligned} & (1-x)\left((-1)^r{ }^{2007} C_r x^{3 r}\right) \\ & (-1)^{r 2007} C_r x^{3 r}-(-1)^{r 2007} C_r x^{3 r+1} \\ & 3 r=2012 \\ & r \neq \frac{2012}{3} \\ & 3 r+1=2012 \\ & 3 r=2011 \\ & r \neq \frac{2011}{3} \end{aligned}$
Hence there is no term containing $\mathrm{x}^{2012}$.
So coefficient of $\mathrm{x}^{2012}=0$
The remainder, when $7^{103}$ is divided by 17, is __________
Explanation:
$ \begin{aligned} & =7 \times(49)^{51} \\\\ & =7 \times(51-2)^{51} \end{aligned} $
Remainder = $7 \times(-2)^{51}$
$ \begin{aligned} & =-7\left(2^3 \cdot(16)^{12}\right) \\\\ & =-56(17-1)^{12} \end{aligned} $
Remainder $=-56 \times(-1)^{12}=-56+68=12$
Let $\alpha$ be the constant term in the binomial expansion of $\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)^{n}, n \leq 15$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{-n}$ is $\lambda \alpha$, then $\lambda$ is equal to _____________.
Explanation:
The $r$-th term of the binomial expansion of $(a+b)^n$ is given by
$T_{r} = {}^n{C_r}a^{n-r}b^{r}$.
Substitute $a$ and $b$ in this formula, we get:
$T_{r} = {}^n{C_r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = {}^n{C_r}(-6)^r x^{\frac{n-4r}{2}}$.
The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.
This happens when $\frac{n-4r}{2} = 0$, which gives $n = 4r$.
$ \begin{aligned} & { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\\\ & (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \\\\ & \text { By observation (625 + 24 = 649) , we get n = 4 } \\\\ & \therefore \alpha=-24 \end{aligned} $
Now, for coefficient of $x^{-4}$
$ \begin{aligned} & \frac{n-4 r}{2}=-4 \\\\ & n=4 r-8 \Rightarrow r=3 \\\\ & \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\\\ & \Rightarrow \lambda=36 \end{aligned} $
The mean of the coefficients of $x, x^{2}, \ldots, x^{7}$ in the binomial expansion of $(2+x)^{9}$ is ___________.
Explanation:
$ T_{r+1}={ }^n C_r 2^{n-r} \times x^r $
Coefficient of $x\left(T_1\right)={ }^9 C_1 \times 2^8$
Coefficient of $x^2\left(T_2\right)={ }^9 C_2 \times 2^7$
Coefficient of $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
              .                    .
              .                    .
              .                    .
Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$
$ \text { Mean }=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7} $
$ \begin{aligned} & { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots .+{ }^9 C_7 \times 2^2 \\ & =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7} \end{aligned} $
$ =\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736 $
The number of integral terms in the expansion of $\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$ is equal to ___________.
Explanation:
The term will be integral if $r$ is a multiple of 4 .
$ \begin{gathered} \therefore r=0,4,8,12, \ldots, 680(\text { which is an } \mathrm{AP}) \\\\ 680=0+(n-1) 4 \\\\ n=\frac{680}{4}+1=171 \end{gathered} $
The coefficient of $x^7$ in ${(1 - x + 2{x^3})^{10}}$ is ___________.
Explanation:
So, general term is $\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$
Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
Now, for possibility,
$\begin{array}{ccc}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 & 1 & 2 \\ 5 & 4 & 1\end{array}$
Thus, required co-efficient
$ \begin{aligned} & =\frac{10 !}{3 ! 7 !}(-1)^7+\frac{10 !}{5 ! 4 !}(-1)^4(2)+\frac{10 !}{7 ! 2 !}(-1)^1(2)^2 \\\\ & =-120+2520-1440 \\\\ & =2520-1560=960 \end{aligned} $
Let $[t]$ denote the greatest integer $\leq t$. If the constant term in the expansion of $\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7}$ is $\alpha$, then $[\alpha]$ is equal to ___________.
Explanation:
$ \mathrm{T}_{r+1}={ }^7 \mathrm{C}_r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r $
For constant term, power of $x$ should be zero.
$ \begin{aligned} & \text { i.e., } 14-2 r-5 r=0 \\\\ & \Rightarrow 14=7 r \Rightarrow r=2 \end{aligned} $
Now, constant term $=\alpha$
$ \begin{aligned} & \Rightarrow{ }^7 C_2(3)^5\left(\frac{-1}{2}\right)^2=\alpha \\\\ & \Rightarrow 21 \times 243 \times \frac{1}{4}=\alpha \\\\ & \Rightarrow[\alpha]=[1275.75]=1275 \end{aligned} $
The largest natural number $n$ such that $3^{n}$ divides $66 !$ is ___________.
Explanation:
$ \begin{aligned} & {\left[\frac{66}{3}\right]=22} \\\\ & {\left[\frac{66}{3^2}\right]=7} \\\\ & {\left[\frac{66}{3^3}\right]=2} \end{aligned} $
Highest powers of 3 is greater than 66. So, their g.i.f. is always 0.
$\therefore$ Required natural number $=22+7+2=31$
The coefficient of $x^{18}$ in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ is __________.
Explanation:
$\begin{aligned} \therefore 60-7 r =18 \\\\ \Rightarrow 7 r =42 \\\\ \Rightarrow r =6\end{aligned}$
$\therefore$ The coefficient of $x^{18}$
$ ={ }^{15} C_6(-1)^6=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=5005 $
Let the sixth term in the binomial expansion of ${\left( {\sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}} + \root 5 \of {{2^{(x - 2){{\log }_2}3}}} } \right)^m}$ in the increasing powers of $2^{(x-2) \log _{2} 3}$, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $x$ is __________.
Explanation:
$ \begin{aligned} \therefore \quad & a={ }^m C_1 \\\\ & a+2 d={ }^m C_2 \\\\ & a+4 d={ }^m C_3 \\\\ \therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\ \Rightarrow & m=7 \text { or } m=2 \\\\ \because & m=2 \text { is not possible } \\\\ \therefore & m=7 \end{aligned} $
$\mathrm{T}_6={ }^{\mathrm{m}} \mathrm{C}_5\left(10-3^{\mathrm{x}}\right)^{\frac{\mathrm{m}-5}{2}} \cdot\left(3^{\mathrm{x}-2}\right)=21$
Putting value of m = 7, we get
$\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\\\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1\end{aligned}$
$ \begin{aligned} & \Rightarrow \left(3^x\right)^2-10 \cdot 3^x+9=0 \\\\ & \Rightarrow 3^x=9,1 \\\\ & \Rightarrow x=0,2 \end{aligned} $
Sum of squares of values of x = 02 + 22 = 4
If the term without $x$ in the expansion of $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}$ is 7315 , then $|\alpha|$ is equal to ___________.
Explanation:
$ T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r $
For constant term
$ \begin{aligned} & \frac{44-2 r}{3}-3 r=0 \\\\ & \Rightarrow r=4 \end{aligned} $
Now ${ }^{22} \mathrm{C}_4 \alpha^4=7315$
$ \begin{aligned} & \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\ & \therefore \alpha^4=1 \\\\ & \therefore |\alpha|=1 \end{aligned} $
The remainder, when $19^{200}+23^{200}$ is divided by 49 , is ___________.
Explanation:
= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
$=49 \lambda+470$
$=49(\lambda+9)+29$
$ \therefore $ Remainder $=29$
expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is
Explanation:
$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^{2}}\right)^{r} \\\\ 9-3 r & =-6 \\\\ r & =5 \end{aligned} $
Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{5}{2}\right)^{5}$ $ =5040 $
Explanation:
$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\ & ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r} \end{aligned} $
For constant term, power of x is zero.
So, $\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45$
Now constant term $=-84$
and ${ }^{9} C_{r} \cdot 2^{3 r-9}(-1)^{r}=-84$
So, $r=3$ and $l=5$
Now for $x^{-15}, \frac{45-5 r}{2}-5 r=-15$
$ \Rightarrow $ $ 45-15 r=-30 $
$ \Rightarrow $ $ r=5 $
$\therefore $ Coefficient $=-{ }^{9} C_{5} 2^{6}=-63.2^{7}$
$\therefore \alpha=7, \beta=-63$
and $|\alpha l-\beta|=|7 \times 5+63|=98$
The remainder on dividing $5^{99}$ by 11 is ____________.
Explanation:
$=625\left[5^{5}\right]^{19}$
$=625[3125]^{19}$
$=625[3124+1]^{19}$
$=625[11 \mathrm{k} \times 19+1]$
$=625 \times 11 \mathrm{k} \times 19+625$
$=11 \mathrm{k}_{1}+616+9$
$=11\left(\mathrm{k}_{2}\right)+9$
Remainder $=9$
Let $\alpha>0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$ has a term $\beta x^{-\alpha}, \beta \in \mathbb{N}$. Then $\alpha$ is equal to ___________.
Explanation:
$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
$\frac{60-11 \mathrm{r}}{3}<0 $
$\Rightarrow 11 \mathrm{r}>60 $
$\Rightarrow \mathrm{r}>\frac{60}{11} $
$\Rightarrow \mathrm{r}=6$
$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
$\therefore \alpha=2$
Explanation:
Given ${x^{{1 \over {50}}}} = 12 \Rightarrow x = {12^{50}}$
${y^{{1 \over {50}}}} = 18 \Rightarrow y = {18^{50}}$
$12\equiv13$ (Mod 25)
$12^2\equiv19$ (Mod 25)
$12^3\equiv-3$ (Mod 25)
$12^9\equiv-2$ (Mod 25)
$12^{10}\equiv-1$ (Mod 25)
$12^{50}\equiv-1$ (Mod 25) ..... (i)
Now
$18\equiv7$ (Mod 25)
$18^2\equiv-1$ (Mod 25)
$18^{-50}\equiv-1$ (Mod 25) ..... (ii)
$\therefore$ $12^{50}+18^{50}\equiv-2$ (Mod 25)
$\equiv23$ (Mod 25)
$\therefore$ Answer = 23
Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.
Explanation:
$ \begin{aligned} & \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\ & \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\ & \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\ & \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\ & \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2) \end{aligned} $
From (1) and (2)
$ \Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8 $
$(1) \Rightarrow r=4$
so, coefficient of middle term is
$ { }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120 $
If the co-efficient of $x^9$ in ${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$ and the co-efficient of $x^{-9}$ in ${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$ are equal, then $(\alpha\beta)^2$ is equal to ___________.
Explanation:
$\because$ Both are equal
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
$\Rightarrow \frac{1}{\beta}=-\alpha$
$\Rightarrow \alpha \beta=-1$
$\Rightarrow(\alpha \beta)^{2}=1$
The remainder when (2023)$^{2023}$ is divided by 35 is __________.
Explanation:
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7
The constant term in the expansion of ${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$ is ___________.
Explanation:
$ \begin{aligned} & \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5} \end{aligned} $
Term independent of $x=$ coefficient of $x^{35}$ in
$ \begin{aligned} & ^{5} C_{4}\left(x^{8}(3 x+2)\right)^{4} \\\\ = & { }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4} \\\\ = & { }^{5} C_{4} \times{ }^{4} C_{3}(2)^{1}(3)^{3} \\\\ = & 5 \times 4 \times 2 \times 27 \\\\ = & 1080 \end{aligned} $
Let the sum of the coefficients of the first three terms in the expansion of ${\left( {x - {3 \over {{x^2}}}} \right)^n},x \ne 0.~n \in \mathbb{N}$, be 376. Then the coefficient of $x^4$ is __________.
Explanation:
$ \begin{aligned} & 3 n^{2}-5 n-250=0 \\\\ & n=10, \frac{-25}{3} \text { (Rejected) } \\\\ & T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\ & ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\ & ={ }^{10} C_{r} x^{10-3 r}(-3)^{r} \end{aligned} $
Here $r=2$
$ \begin{aligned} \text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\ & =45 \times 9 \\\\ & =405 \end{aligned} $
Suppose $\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $. Then the value of $\alpha$ is ___________
Explanation:
Concept :
(1) ${}^n{C_r} = {n \over r}\,.\,{}^{n - 1}{C_{r - 1}}$
Given,
$\sum\limits_{r = 0}^{2023} {{r^2}\,.\,{}^{2023}{C_r}} $
$ = \sum\limits_{r = 0}^{2023} {{r^2}\,.\,{{2023} \over r}\,.{}^{2022}{C_{r - 1}}} $
$ = 2023\sum\limits_{r = 0}^{2023} {{r}\,.\,{}^{2022}{C_{r - 1}}} $
$ = 2023\sum\limits_{r = 0}^{2023} {[(r - 1) + 1]\,.\,{}^{2022}{C_{r - 1}}} $
$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}} + \sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}} } } \right]$
$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{{2022} \over {(r - 1)}}\,.\,{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$
$ = 2023\left[ {2022\sum\limits_{r = 0}^{2023} {{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$
$ = 2023\left[ {2022\,.\,{2^{2021}} + {2^{2022}}} \right]$
$ = 2023\,.\,{2^{2021}}\left[ {2022 + 2} \right]$
$ = 2023\,.\,{2^{2021}}\,.\,2024$
$ = 2023\,.\,{{{2^{2022}}} \over {{2}}}\,.\,2024$
$ = 2023\,.\,{2^{2022}}\,.\,{1012}$
$\therefore$ $\alpha = {1012}$
$ \text { If } \sum\limits_{k=1}^{10} K^{2}\left(10_{C_{K}}\right)^{2}=22000 L \text {, then } L \text { is equal to }$ ________.
Explanation:
Given,
$\sum\limits_{k = 1}^{10} {{k^2}{{\left( {{}^{10}{C_k}} \right)}^2} = 2200\,L} $
$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{}^{10}{C_k}} \right)}^2} = 22000\,L} $
$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{{10} \over k}\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $
$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {10\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $
$ \Rightarrow 100\,.\,\sum\limits_{k = 1}^{10} {{{\left( {{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $
$ \Rightarrow 100\left( {{{\left( {{}^9{C_0}} \right)}^2} + {{\left( {{}^9{C_1}} \right)}^2}\, + \,....\, + \,{{\left( {{}^9{C_9}} \right)}^2}} \right) = 22000\,L$
$ \Rightarrow 100\left( {{}^{18}{C_9}} \right) = 22000\,L$
[Note : ${\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2}\, + \,....\, + \,{\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$]
$ \Rightarrow 100 \times {{18!} \over {9!\,9!}} = 22000\,L$
$ \Rightarrow L = 221$
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$, in the increasing powers of $\frac{1}{\sqrt[4]{3}}$ be $\sqrt[4]{6}: 1$. If the sixth term from the beginning is $\frac{\alpha}{\sqrt[4]{3}}$, then $\alpha$ is equal to _________.
Explanation:
Fifth term from beginning $ = {}^n{C_4}{\left( {{2^{{1 \over 4}}}} \right)^{n - 4}}{\left( {{3^{{{ - 1} \over 4}}}} \right)^4}$
Fifth term from end $ = {(n - 5 + 1)^{th}}$ term from begin $ = {}^n{C_{n - 4}}{\left( {{2^{{1 \over 4}}}} \right)^3}{\left( {{3^{{{ - 1} \over 4}}}} \right)^{n - 4}}$
Given ${{{}^n{C_4}{2^{{{n - 4} \over 4}}}\,.\,{3^{ - 1}}} \over {{}^n{C_{n - 3}}{2^{{4 \over 4}}}\,.\,{3^{ - \left( {{{n - 4} \over 4}} \right)}}}} = {6^{{1 \over 4}}}$
$ \Rightarrow {6^{{{n - 8} \over 4}}} = {6^{{1 \over 4}}}$
$ \Rightarrow {{n - 8} \over 4} = {1 \over 4} \Rightarrow n = 9$
${T_6} = {T_{5 + 1}} = {}^9{C_5}{\left( {{2^{{1 \over 4}}}} \right)^4}{\left( {{3^{{{ - 1} \over 4}}}} \right)^5}$
$ = {{{}^9{C_5}\,.\,2} \over {{3^{{1 \over 4}}}\,.\,3}} = {{84} \over {{3^{{1 \over 4}}}}} = {\alpha \over {{3^{{1 \over 4}}}}}$
$ \Rightarrow \alpha = 84.$
Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an A.P. If d is the common difference of this A.P. , then $50-\frac{2 d}{\beta^{2}}$ is equal to __________.
Explanation:
Coefficients of middle terms of given expansions are ${}^4{C_2}{1 \over 6}{\beta ^2},\,{}^2{C_1}( - 3\beta ),\,{}^6{C_3}{\left( {{{ - \beta } \over 2}} \right)^3}$ form an A.P.
$\therefore$ $2.2( - 3\beta ) = {\beta ^2} - {{5{\beta ^3}} \over 2}$
$ \Rightarrow - 24 = 2\beta - 5{\beta ^2}$
$ \Rightarrow 5{\beta ^2} - 2\beta - 24 = 0$
$ \Rightarrow 5{\beta ^2} - 12\beta + 10\beta - 24 = 0$
$ \Rightarrow \beta (5\beta - 12) + 2(5\beta - 12) = 0$
$\beta = {{12} \over 5}$
$d = - 6\beta - {\beta ^2}$
$\therefore$ $50 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57$
If $1 + (2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}})$ is equal to $2^{\mathrm{n}} \cdot \mathrm{m}$, where $\mathrm{m}$ is odd, then $\mathrm{n}+\mathrm{m}$ is equal to __________.
Explanation:
$l = 1 + (1 + {}^{49}{C_0} + {}^{49}{C_1}\, + \,....\, + \,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}})$
As ${}^{49}{C_0} + {}^{49}{C_1}\, + \,.....\, + \,{}^{49}{C_{49}} = {2^{49}}$
and ${}^{50}{C_0} + {}^{50}{C_2}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}}$
$ \Rightarrow {}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}} - 1$
$\therefore$ $l = 1 + ({2^{49}} + 1)({2^{49}} - 1)$
$ = {2^{98}}$
$\therefore$ $m = 1$ and $n = 98$
$m + n = 99$
Let for the $9^{\text {th }}$ term in the binomial expansion of $(3+6 x)^{\mathrm{n}}$, in the increasing powers of $6 x$, to be the greatest for $x=\frac{3}{2}$, the least value of $\mathrm{n}$ is $\mathrm{n}_{0}$. If $\mathrm{k}$ is the ratio of the coefficient of $x^{6}$ to the coefficient of $x^{3}$, then $\mathrm{k}+\mathrm{n}_{0}$ is equal to :
Explanation:
${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$
If T9 is numerically greatest term
$\therefore$ ${T_8} \le {T_9} \le {T_{10}}$
${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$
$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {(n - 8)!8!}}3\,.\,(6x) \ge {{n!} \over {(n - 9)!9!}}{(6x)^2}$
$ \Rightarrow \underbrace {{9 \over {(n - 7)(n - 8)}}}_{} \le \underbrace {{{18\left( {{3 \over 2}} \right)} \over {(n - 8)8}} \ge {{36} \over {9.8}}{9 \over 4}}_{}$
$72 \le 27(n - 7)$ and $27 \ge 9(n - 8)$
${{29} \over 3} \le n$and $n \le 11$
$\therefore$ ${n_0} = 10$
For ${(3 + 6x)^{10}}$
${T_{r + 1}} = {}^{10}{C_r}$
${3^{10 - r}}{(6x)^r}$
For coeff. of x6
$r = 6 \Rightarrow {}^{10}{C_6}{3^4}{.6^6}$
For coeff. of x3
$r = 3 \Rightarrow {}^{10}{C_3}{3^7}{.6^3}$
$\therefore$ $k = {{{}^{10}{C_6}} \over {{}^{10}{C_3}}}.{{{3^4}{{.6}^6}} \over {{3^7}{{.6}^3}}} = {{10!7!3!} \over {6!4!10!}}.8$
$ \Rightarrow k = 14$
$\therefore$ $k + {n_0} = 24$
If the coefficients of $x$ and $x^{2}$ in the expansion of $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}, \mathrm{p}, \mathrm{q} \leq 15$, are $-3$ and $-5$ respectively, then the coefficient of $x^{3}$ is equal to _____________.
Explanation:
Coefficient of x in ${(1 + x)^p}{(1 - x)^q}$
$ - {}^p{C_0}\,{}^q{C_1} + {}^p{C_1}\,{}^q{C_0} = - 3 \Rightarrow p - q = - 3$
Coefficient of x2 in ${(1 + x)^p}{(1 - x)^q}$
${}^p{C_0}\,{}^q{C_2} - {}^p{C_1}\,{}^q{C_1} + {}^p{C_2}\,{}^q{C_0} = - 5$
${{q(q - 1)} \over 2} - pq + {{p(q - 1)} \over 2} = - 5$
${{{q^2} - q} \over 2} - (q - 3)q + {{(q - 3)(q - 4)} \over 2} = - 5$
$ \Rightarrow q = 11,\,p = 8$
Coefficient of x3 in ${(1 + x)^8}{(1 - x)^{11}}$
$ = - {}^{11}{C_3} + {}^8{C_1}\,{}^{11}{C_2} - {}^8{C_2}\,{}^{11}{C_1} + {}^8{C_3} = 23$