JEE Main 2014 — Mathematics Algebra

Given ${(1+{x}^{n}+{x}^{253})}^{10}$(1+xn+x253)10

$$={{(1+{x}^{253})+{x}^{n}}}^{10}$$=(1+x253)+xn10

Using the binomial expansion ${(a+b)}^{n}=C0n{a}^{n}{b}^{0}+C1n{a}^{n-1}b+C2n{a}^{n-2}{b}^{2}+...+Cnn{a}^{0}{b}^{n},$(a+b)n=C0nanb0+C1nan−1b+C2nan−2b2+...+Cnna0bn,

$$={ }^{10}{C}_{0}{(1+{x}^{253})}^{10}{({x}^{n})}^{0}+{ }^{10}{C}_{1}{(1+{x}^{253})}^{9}{({x}^{n})}^{1}+{ }^{10}{C}_{2}{(1+{x}^{253})}^{8}{({x}^{n})}^{2}+...+{ }^{10}{C}_{10}{(1+{x}^{253})}^{0}{({x}^{n})}^{10}$$=10C0​(1+x253)10(xn)0+10C1​(1+x253)9(xn)1+10C2​(1+x253)8(xn)2+...+10C10​(1+x253)0(xn)10

As $2 5 3 = 2 3 \times 1 1$253=23×11 and $1012=253\times 4$1012=253×4, also $n\leq 22$n≤22

$\Rightarrow$⇒Coefficient of ${x}^{1012}$x1012 will come only from the first term, i.e. in

$${ }^{10}{C}_{0}{(1+{x}^{253})}^{10}{({x}^{n})}^{0}={(1+{x}^{253})}^{10}$$10C0​(1+x253)10(xn)0=(1+x253)10

The general term in the expansion of ${(1+a)}^{n}$(1+a)n is ${T}_{r+1}=Crn{a}^{r}$Tr+1​=Crnar

Hence, the general term in the expansion of ${(1+{x}^{253})}^{10}$(1+x253)10 is ${T}_{r+1}=Cr10{({x}^{253})}^{r}=Cr10({x}^{253r})$Tr+1​=Cr10(x253)r=Cr10(x253r)

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

Thus, the required coefficient is$={ }^{10}{C}_{4}.$=10C4​.