Binomial Theorem
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 $\mathrm{x}^{2012}=0$
If $p_{1}=20$ and $p_{2}=210$, then $2(a+b+c)$ is equal to :
The coefficient of $x^{5}$ in the expansion of $\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$ is :
Fractional part of the number $\frac{4^{2022}}{15}$ is equal to
If $\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$ then $n$ is equal to :
The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1-x)^{100}$, is equal to
The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}+2}$, which are in the ratio $1: 3: 5$, is equal to :
If the $1011^{\text {th }}$ term from the end in the binominal expansion of $\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{2022}$ is 1024 times $1011^{\text {th }}$R term from the beginning, then $|x|$ is equal to
Let the number $(22)^{2022}+(2022)^{22}$ leave the remainder $\alpha$ when divided by 3 and $\beta$ when divided by 7. Then $\left(\alpha^{2}+\beta^{2}\right)$ is equal to :
If the coefficients of $x$ and $x^{2}$ in $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$ are 4 and $-$5 respectively, then $2 p+3 q$ is equal to :
If the coefficient of ${x^7}$ in ${\left( {ax - {1 \over {b{x^2}}}} \right)^{13}}$ and the coefficient of ${x^{ - 5}}$ in ${\left( {ax + {1 \over {b{x^2}}}} \right)^{13}}$ are equal, then ${a^4}{b^4}$ is equal to :
$25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by :
The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of $\left(2 x^{2}+\frac{1}{2 x}\right)^{11}$ is equal to :
If the coefficients of three consecutive terms in the expansion of $(1+x)^{n}$ are in the ratio $1: 5: 20$, then the coefficient of the fourth term is
If the coefficient of ${x^7}$ in ${\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}}$ and ${x^{ - 7}}$ in ${\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}}$ are equal, then :
Among the statements :
(S1) : $2023^{2022}-1999^{2022}$ is divisible by 8
(S2) : $13(13)^{n}-12 n-13$ is divisible by 144 for infinitely many $n \in \mathbb{N}$
If ${ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1$, then the ratio $\left(n^{2}+3 n\right):\left(n^{2}-3 n+4\right)$ is :
If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$ is $\sqrt{6}: 1$, then the third term from the beginning is :
If the coefficient of $x^{15}$ in the expansion of $\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15}$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $(\mathrm{a}, \mathrm{b})$ :
The coefficient of ${x^{301}}$ in ${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$ is :
Let K be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+x)^{99}$. Let $a$ be the middle term in the expansion of ${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}$. If ${{{}^{200}{C_{99}}K} \over a} = {{{2^l}m} \over n}$, where m and n are odd numbers, then the ordered pair $(l,\mathrm{n})$ is equal to
If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1 + x)^{10}$, then $\sum\limits_{r = 1}^{10} {{r^3}{{\left( {{{{a_r}} \over {{a_{r - 1}}}}} \right)}^2}} $ is equal to
If ${({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}$ then $\alpha$ is equal to :
The value of $\sum\limits_{r = 0}^{22} {{}^{22}{C_r}{}^{23}{C_r}} $ is
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 $