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 $x^{1012}$
$=\frac{10!}{6!0!4!}$
$={}^{10}{C}_4$