Question

If number of terms in the expansion of ${(1+4x+x^2)}^n,n\in N$ is $1225$, then the value of $n$ is-

• A. $48$
• B. $408$
• C. $612$
• D. $1244$

Answer 1

Answer: (A)

$48$

We know that number of terms in expansion of $(x_1+x_2+x_3+.....+x_r{)}^n$ is ${}^{n+r-1}{C}_n$.

Hence number of terms in $(1+4x+x^2{)}^n$ is ${}^{n+3-1}{C}_n{=}^{n+2}{C}_2=1225$ (given)

$\Rightarrow \frac{(n+2)(n+1)}{2}=1225\Rightarrow (n+2)(n+1)=2450$

$\Rightarrow n^2+3n-2448=0\Rightarrow n=48$