The coefficient of x^{1012} in the expansion | vedclass

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Question

Given expansion $\left(1+x^{n}+x^{253}\right)^{10}$

Let $x^{1012}=(1)^{a}\left(x^{n}\right)^{b} \cdot\left(x^{253}\right)^{c}$

Here $a, b, c, n$

Vedclass · Accepted Answer

$^{10}{C_4}$

Vedclass · Answer

Given expansion $\left(1+x^{n}+x^{253}ight)^{10}$

Let $x^{1012}=(1)^{a}\left(x^{n}ight)^{b} \cdot\left(x^{253}ight)^{c}$

Here $a, b, c, n$ are all $+ve$ integers and

$a$ $\leq 10, b \leq 10, c \leq 4, n \leq 22, a+b+c=10$

Now $b n+253 c=1012$

$\Rightarrow b n=253(4-c)$

For $c<4$ and $n \leq 22 ; b>10,$ which is not possible.

$	herefore c=4, b=0, a=6$

$	herefore x^{1012}=(1)^{6} \cdot\left(x^{12}ight)^{0} \cdot\left(x^{253}ight)^{4}$

Hence the coefficient of $x^{1012}$

$=\frac{10 !}{6 ! 0 ! 4 !}=10 \mathrm{C}_{4}$

The coefficient of $x^{1012}$ in the expansion of ${\left( {1 + {x^n} + {x^{253}}} \right)^{10}}$ , (where $n \leq 22$ is any positive integer), is

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