Binomial Theorem

$ \begin{aligned} & \text { }{ }^n C_{n-r}+3 \cdot{ }^n C_{n-r+1}+3 \cdot{ }^n C_{n-r+2}+{ }^n C_{n-r+3}={ }^x C_r \\ & \Rightarrow\left({ }^n C_{n-r}+{ }^n C_{n-r+1}\right)+2 \cdot\left({ }^n C_{n-r+1}+{ }^n C_{n-r+2}\right) +\left({ }^n C_{n-r+2}+{ }^n C_{n-r+3}\right)={ }^n C_r \end{aligned} $

$ \begin{aligned} & \Rightarrow{ }^{n+1} C_{n-r+1}+2^{n+1} C_{n-r+2}+{ }^{n+1} C_{n-r+3}={ }^x C_r \\ & {\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right] } \end{aligned} $

$ \begin{aligned} \Rightarrow\left\{{ }^{n+1} C_{n-r+1}+{ }^{n+1} C_{n-r+2}\right\} +\left\{{ }^{n+1} C_{n-r+2}+{ }^{n+1} C_{n-r+3}\right\}={ }^x C_r \end{aligned} $

$ \begin{aligned} & \Rightarrow{ }^{n+2} C_{n-r+2}+{ }^{n+2} C_{n-r+3}={ }^x C_r \\ & \Rightarrow{ }^{n+3} C_{n-r+3}={ }^x C_r \end{aligned} $

$ \begin{aligned} &\Rightarrow{ }^{n+3} C_r={ }^x C_r\,\,\,\,\,\,\left[\because{ }^{n+3} C_{n-r+3}={ }^{n+3} C_r\right] \end{aligned} $

$ \therefore \quad x=n+3 $

We have, $(1+x)^{41}\left(1-x+x^2\right)^{40}$

$ \begin{aligned} & =\left\{(1+x)\left(1-x+x^2\right)\right\}^{40}(1+x) \\ & =\left(1+x^3\right)^{40}(1+x) \\ & =\left(1+x^3\right)^{40}+x\left(1+x^3\right)^{40} \end{aligned} $

∴ Coefficient of $x^{50}$ in $(1+x)^{41}\left(1-x+x^2\right)^{40}$

= Coefficient of $x^{50}$ in $\left(1+x^3\right)^{40}+$ Coefficient of $x^{50}$ in $x\left(1+x^3\right)^{40}$

$=$ Coefficient of $x^{50}$ in $\left(1+x^3\right)^{40}+$ Coefficient of $x^{49}$ in $\left(1+x^3\right)^{40}$

= Coefficient of $x^{50}$ in $\sum_{r=0}^{40}{ }^{40} C_r x^{3 r}+$ Coefficient of $x^{50}$ in $\sum_{r=0}^{40}{ }^{40} C_r x^{3 r}=0+0=0$

General term of given binomial expansion is

$ \begin{aligned} T_{r+1} & ={ }^{10} C_{r}\left(\frac{1}{3} x^{\frac{1}{2}}\right)^{10-r}\left(x^{-\frac{1}{4}}\right)^{r} \\ & ={ }^{10} C_{r} \times\left(\frac{1}{3}\right)^{10-r} x^{\frac{10-r}{2}-\frac{r}{4}} \end{aligned} $

We have to find coefficient of $ x^{2} $.

$ \begin{aligned} \frac{10-r}{2}-\frac{r}{4} & =2 \Rightarrow r=4 \\ \Rightarrow \quad T_{4+1} & ={ }^{10} C_{4}\left(\frac{1}{3}\right)^{6} x^{2} \\ \therefore \text { Coefficient of } x^{2} & ={ }^{10} C_{4}\left(\frac{1}{3}\right)^{6} \\ & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{1}{3^{6}}=\frac{70}{243} \end{aligned} $

We have, $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1$

$\begin{array}{r} =x^5+5 x^4 \sqrt{x^3-1}+10 x^3\left(\sqrt{x^3-1}\right)^2+10 x^2\left(\sqrt{x^3-1}\right)^3 \\ +5 x\left(\sqrt{x^3-1}\right)^4+\left(\sqrt{x^3-1}\right)^5+\left(x^5-5 x^2 \sqrt{x^3-1}+\right. \\ 10 x^3\left(\sqrt{x^3-1}\right)^2-10 x^2\left(\sqrt{x^3-1}\right)^3+5 x\left(\sqrt{x^3-1}\right)^4 \\ -\left(\sqrt{x^3-1}\right)^5 \end{array}$

$\begin{aligned} & =2\left(x^5+10 x^3\left(x^3-1\right)+5 x\left(x^3-1\right)^2\right) \\ & =2\left(x^5+10 x^6-10 x^3+5 x^7-10 x^4+5 x\right) \end{aligned}$

Now, the sum of coefficients of all odd degree term is

$2(1-10+5+5)=2$

$\text { Given, } \sum_\limits{\substack{i, j=0 \\ i \neq j}}^n C_i^n C_j$

$=\sum_\limits{i, j=0}^n{ }^n C_i^n C_j-\sum_\limits{\substack{i, j=0 \\ i=j}}^n{ }^n C_i^n C_j$

$\begin{aligned} & =\left(\sum_{i=0}^n{ }^n C_i\right)\left(\sum_{j=0}^n{ }^n C_j\right)-\sum_{i=0}^n\left({ }^n C_i\right)^2 \\ & =2^n \times 2^n-{ }^{2 n} C_n \end{aligned}$

$\left[\because \sum_{i=0}^n{ }^n C_i=2^n \text { and } \sum_{i=0}^n\left({ }^n C_i\right)^2={ }^{2 n} C_n\right]$

$=2^{2 n}-{ }^{2 n} C_n$

We have,

${(1 + 5\sqrt {2x} )^9} + {(1 - 5\sqrt {2x} )^9}$

$ = 2\{ {}^2{C_0} + {}^9{C_2}{(5\sqrt 2 x)^2}\, + \,...\, + \,{}^9{C_8}{(5\sqrt 2 x)^8}\} $

Clearly, it has 5 terms.

Given, sum of the coefficient = 1024

i.e. 2ⁿ = 1024 = 2¹⁰

$\Rightarrow$ n = 10

Since, n is even, so greatest coefficient

= ⁿC_n/2 = ¹⁰C₅ = 252

${{{C_1}} \over {{C_0}}} = n,{{{C_2}} \over {{C_1}}} = {{(n - 1)} \over 2},{{{C_3}} \over {{C_2}}} = {{n - 2} \over 3}$ and so on.

${{{C_1}} \over {{C_0}}} + 2{{{C_2}} \over {{C_1}}} + 3{{{C_3}} \over {{C_2}}} + ....n{{{C_n}} \over {{C_{n - 1}}}} = \sum {n = {{n(n + 1)} \over 2}} $

On putting n = 20,

${{{C_1}} \over {{C_0}}} + 2{{{C_2}} \over {{C_1}}} + 3{{{C_3}} \over {{C_2}}} + ....20{{{C_{20}}} \over {{C_{19}}}} = {{20 \times 21} \over 2} = 210$

${}^{47}{C_4} + \sum\limits_{r = 1}^5 {{}^{52 - r}{C_3} = {}^{47}{C_4} + {}^{51}{C_3} + {}^{50}{C_3} + {}^{49}{C_3} + {}^{48}{C_3} + {}^{47}{C_3}} $

$ = {}^{51}{C_3} + {}^{50}{C_3} + {}^{49}{C_3} + {}^{48}{C_3} + ({}^{47}{C_3} + {}^{47}{C_4}) = {}^{52}{C_4}$

We have,

$(x - 1)(x - 2)(x - 3)....(x - 10)$

Coefficient of x⁸ = Sum of the terms taken two at a time i.e.,

$\sum\limits_{1 \le i < j \le 10}^{10} {{x_i}\,.\,{x_j} = {1 \over 2}\left[ {{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}^2} - \left( {\sum\limits_{i = 1}^{10} {{x_i}^2} } \right)} \right]} $

$ = {1 \over 2}[{(1 + 2 + 3 + ... + 10)^2} - {({1^2} + {2^2} + {3^2} + ... + 10)^2}]$

$ = {1 \over 2}\left[ {{{\left( {{{10 \times 11} \over 2}} \right)}^2} - \left( {{{10 \times 11 \times 21} \over 6}} \right)} \right]$

$ = {1 \over 2}({(55)^2} - 385) = {1 \over 2}(3025 - 385) = {{2640} \over 2} = 1320$