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)

${^{10}C_4}$

Answer 2

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 $x^{1012}$ $= \frac{10!}{6! 0! 4!}$ $= {^{10}C_4}$