JEE Main 2014 — Binomial Theorem Question with Solution

Given ${\left(1+x^n+x^{253}\right)}^{10}$

$={\left\{\left(1+x^{253}\right)+x^n\right\}}^{10}$

Using the binomial expansion ${\left(a+b\right)}^n={}^{n}C_0{a}^n{b}^0+{}^{n}C_1{a}^{n-1}b+{}^{n}C_2{a}^{n-2}{b}^2+...+{}^{n}C_n{a}^0{b}^n,$

$={}^{10}{C}_0{\left(1+x^{253}\right)}^{10}{\left(x^n\right)}^0+{}^{10}{C}_1{\left(1+x^{253}\right)}^9{\left(x^n\right)}^1+{}^{10}{C}_2{\left(1+x^{253}\right)}^8{\left(x^n\right)}^2+...+{}^{10}{C}_{10}{\left(1+x^{253}\right)}^0{\left(x^n\right)}^{10}$

As $253=23\times 11$ and $1012=253\times 4$, also $n\le 22$

$\Rightarrow$Coefficient of $x^{1012}$ will come only from the first term, i.e. in

${}^{10}{C}_0{\left(1+x^{253}\right)}^{10}{\left(x^n\right)}^0={\left(1+x^{253}\right)}^{10}$

The general term in the expansion of ${\left(1+a\right)}^n$ is $T_{r+1}={}^{n}C_r{a}^r$

Hence, the general term in the expansion of ${\left(1+x^{253}\right)}^{10}$ is $T_{r+1}={}^{10}C_r{\left(x^{253}\right)}^r={}^{10}C_r\left(x^{253r}\right)$

Since, $1012=253\times 4,$ hence $r=4$

Thus, the required coefficient is$={}^{10}{C}_4.$