Question

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

• A. 1
• B. ${^{10}C_4}$
• C. 4n
• D. ${^{253}C_4}$

Answer 1

Answer: (B)

Given expansion $(1+ x^n + x^{253})^{10}$
Let $x^{1012} = (1)^a (x^n)^b. (x^{253})^c$
Here a, b, c, n are all +ve integers and a $\le$ 10, b $\le$ 10, c $\le$ 4, n $\le$ 22, a + b + c = 10
Now $bn + 253c = 1012$
$\Rightarrow \, bn = 253 (4 - c) $
For c < 4 and n $\le$ 22; b > 10, which is not possible.
$\therefore $ c = 4, b = 0, a =6
$ \therefore \, x^{1012} = (1)^6. (x^n)^0 . (x^{253})^{4}$
Hence the coefficient of $x^{1012}$
$ = \frac{10!}{6! 0! 4!}$
$= {^{10}C_4}$