Question

If $\left(1+x+x^{2}\right)^{n} =1+a_{1}x+a_{2}x^{2} +\cdots+a_{2n}x^{2n}$, $2a_{1} -3a_{2} +\cdots-\left(2n+1\right)a_{2n}$ is equal to

• A. n
• B. #NAME?
• C. n + 1
• D. #NAME?

Answer 1

The Correct Option is B

Given,
$\left(1+x+x^{2}\right)^{n}=1+a_{1} x+a_{2} x^{2}+\ldots+ a_{2 n} x^{2 n}$
$\Rightarrow \, x\left(1+x+x^{2}\right)^{n}=x+a_{1} x^{2}+a_{2} x^{3}+ \ldots +a_{2 n} x^{2 n+1}$
On differentiating w.r.t. $x$, we get
$(1+ \left.x+x^{2}\right)^{n}+x \cdot n\left(1+x+x^{2}\right)^{n-1}(1+2 x)$
$=1+2 a_{1} x+3 a_{2} x^{2}+\ldots+a_{2 n} \cdot(2 n+1) x^{2 n}$
On putting $x=-1$, we get
$(1-1+1)^{n}-n(1-1+1)^{n-1}(1-2)$
$=1-2 a_{1}+3 a_{2}+\ldots+a_{2 n}(2 n+1)$
$\Rightarrow \, 1-n(-1)=1-2 a_{1}+3 a_{2}+\ldots+a_{2 n}(2 n+1)$
$\Rightarrow \, 2 a_{1}-3 a_{2} \ldots-(2 n+1) a_{2 n}=-n$