If the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in arithmetic progression, then $n^2-41 n$ is equal to

Question

Aicharya · Accepted Answer

Given that coefficient of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+n)^n$ are in A.P. It means ${ }^n C_9 \cdot{ }^n C_{10} \cdot{ }^n C_{11}$ are in A.P. $\begin{aligned} & 	herefore \quad 2^n C_{10}={ }^n C_9+{ }^n C_{11} \ & 2=\frac{{ }^n C_9}{{ }^n C_{10}}+\frac{{ }^n C_{11}}{{ }^n C_{10}} \ & \Rightarrow \quad 2=\frac{n !}{9 !(n-9) !} 	imes \frac{10 !(n-10) !}{n !} \end{aligned}$ $ +\frac{n !}{11 !(n-11) !} 	imes \frac{10 !(n-10) !}{n !} $ $\begin{aligned} & \Rightarrow \quad 2=\frac{10}{n-9}+\frac{n-10}{11} \ & 2=\frac{110+(n-10)(n-9)}{11(n-9)} \ & 2=\frac{110+n^2-9 n-10 n+90}{11 n-99} \ & 22 n-198=200+n^2-19 n \ & 	herefore \quad n^2-41 n=-398\end{aligned}$

If the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in arithmetic…

Solution