Binomial Theorem
Given below are two statements :
Statement I :
$25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by 7.
Statement II :
The integral part of $(7 + 4\sqrt{3})^{25}$ is an odd number.
In the light of the above statements, choose the correct answer from the options given below :
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Both Statement I and Statement II are true
Statement I is true but Statement II is false
The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \ldots + x^{1000}$ is :
Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^k}{n!}, k \in \mathrm{~N}$, then $n+k$ is equal to
50
52
49
51
The sum of all possible values of $\mathbf{n} \in \mathbf{N}$, so that the coefficients of $x, x^2$ and $x^3$ in the expansion of $\left(1+x^2\right)^2(1+x)^{\mathrm{n}}$, are in arithmetic progression is :
12
9
3
7
The value of $\frac{{ }^{100} \mathrm{C}_{50}}{51}+\frac{{ }^{100} \mathrm{C}_{51}}{52}+\ldots .+\frac{{ }^{100} \mathrm{C}_{100}}{101}$ is:
$\frac{2^{101}}{101}$
$\frac{2^{100}}{101}$
$\frac{2^{100}}{100}$
$\frac{2^{101}}{100}$
Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If
$P_n=C_0-C_1+\frac{2^2}{3} C_2-\frac{2^3}{4} C_3+\ldots . .+\frac{(-2)^n}{n+1} C_n$, then the value of $\sum\limits_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.
675
580
525
650
The coefficient of $x^{48}$ in $(1+x)+2(1+x)^2+3(1+x)^3+\ldots+100(1+x)^{100}$ is equal to
$100 \cdot{ }^{100} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{48}$
$100 \cdot{ }^{101} \mathrm{C}_{49}-{ }^{101} \mathrm{C}_{50}$
${ }^{100} \mathrm{C}_{50}+{ }^{101} \mathrm{C}_{49}$
$100 \cdot{ }^{100} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{50}$
If the coefficient of $x$ in the expansion of $\left(a x^2+b x+c\right)(1-2 x)^{26}$ is -56 and the coefficients of $x^2$ and $x^3$ are both zero, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to :
1483
1300
1500
1403
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} $
The number of integral terms in the expansion of $ \left( {5^\frac{1}{2}} + 7^\frac{1}{8} \right)^{1016} $ is:
127
128
130
129
The remainder when $\left((64)^{(64)}\right)^{(64)}$ is divided by 7 is equal to
If $1^2 \cdot\left({ }^{15} C_1\right)+2^2 \cdot\left({ }^{15} C_2\right)+3^2 \cdot\left({ }^{15} C_3\right)+\ldots+15^2 \cdot\left({ }^{15} C_{15}\right)=2^m \cdot 3^n \cdot 5^k$, where $m, n, k \in \mathbf{N}$, then $\mathrm{m}+\mathrm{n}+\mathrm{k}$ is equal to :
For an integer $n \geq 2$, if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2 n-3}$ is 16 , then the distance of the point $\mathrm{P}\left(2 n-1, n^2-4 n\right)$ from the line $x+y=8$ is
In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, n \in \mathrm{~N}$, if the ratio of $15^{\text {th }}$ term from the beginning to the $15^{\text {th }}$ term from the end is $\frac{1}{6}$, then the value of ${ }^n \mathrm{C}_3$ is
If $\sum\limits_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r=\alpha\left(\frac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathbb{N}$, then $(\alpha+\beta)^2$ is equal to
The largest $\mathrm{n} \in \mathbf{N}$ such that $3^{\mathrm{n}}$ divides 50 ! is :
The term independent of $x$ in the expansion of $\left(\frac{(x+1)}{\left(x^{2 / 3}+1-x^{1 / 3}\right)}-\frac{(x-1)}{\left(x-x^{1 / 2}\right)}\right)^{10}, x>1$, is :
The remainder, when $7^{103}$ is divided by 23, is equal to:
9
6
14
17
The least value of n for which the number of integral terms in the Binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^n$ is 183, is :
2184
2172
2196
2148
Let the coefficients of three consecutive terms $T_r$, $T_{r+1}$ and $T_{r+2}$ in the binomial expansion of $(a + b)^{12}$ be in a G.P. and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $p + q$ is equal to:
295
283
299
287
Suppose $A$ and $B$ are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 \mathrm{n}-1}$. If $2 \mathrm{~A}=5 \mathrm{~B}$, then n is equal to:
For some $\mathrm{n} \neq 10$, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+\mathrm{x})^{\mathrm{n}+4}$ is:
If in the expansion of $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$, the coefficients of $x$ and $x^2$ are 1 and -2 , respectively, then $\mathrm{p}^2+\mathrm{q}^2$ is equal to :
Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of
$\begin{aligned}
& \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1 \text {. If } u \text { and } v \text { satisfy the equations } \\\\
& \alpha u+\beta v=18, \\\\
& \gamma u+\delta v=20,
\end{aligned}$
then $\mathrm{u+v}$ equals :
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 sum of the coefficient of $x^{2 / 3}$ and $x^{-2 / 5}$ in the binomial expansion of $\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9$ is
The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$. Then a possible value of $\mathrm{p}+\mathrm{q}$ is :
If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$ is 105 , then $\mathrm{a}^2$ is equal to :
If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25 \alpha$ is equal to :
If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in the arithmetic progression, then the maximum value of $n$ is:
The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :
in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is :
Let $a$ be the sum of all coefficients in the expansion of $\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}$ and $b=\lim _\limits{x \rightarrow 0}\left(\frac{\int_0^x \frac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)$. If the equation $c x^2+d x+e=0$ and $2 b x^2+a x+4=0$ have a common root, where $c, d, e \in \mathbb{R}$, then $\mathrm{d}: \mathrm{c}:$ e equals
Suppose $2-p, p, 2-\alpha, \alpha$ are the coefficients of four consecutive terms in the expansion of $(1+x)^n$. Then the value of $p^2-\alpha^2+6 \alpha+2 p$ equals
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:
$\begin{aligned} & (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\ & (x+2)^2+\ldots \ldots .+(x+2)^{n-1} \\ & \sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} \\ & =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots .+\left(\frac{3}{4}\right)^{n-1}\right] \\ & =4^{n-1} \times \frac{1-\left(\frac{3}{4}\right)^n}{1-\frac{3}{4}} \\ & =4^n-3^n=\beta^n-\gamma^n \\ & \beta=4, \gamma=3 \\ & \beta^2+\gamma^2=16+9=25 \end{aligned}$