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If $a_k$ is the coefficient of $x^k$ in the expansion of
$\left(1+x+x^2\right)^n$ for $k=0,1,2, \ldots, 2 n$, then $a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n}$ is equal to
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$\left(1+x+x^2\right)^n$ for $k=0,1,2, \ldots, 2 n$, then $a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n}$ is equal to
Solution:
2909 Upvotes
Verified Answer
The correct answer is:
$n \cdot 3^n$
We have,
$\begin{aligned}\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+a_3 x^3 & +\ldots \\ & +a_{2 \pi} x^{2 n}\end{aligned}$
On differentiating both sides, we get
$\begin{array}{r}n\left(1+x+x^2\right)^{n-1}(1+2 x)=a_1+2 a_2 x+3 a_3 x \\ +\ldots \ldots+2 n a_{2 n} x^{2 n-1}\end{array}$
Now, putting $x=1$, we get
$\begin{aligned} & n(3)^{n-1} \cdot(3)=a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n} \\ & \Rightarrow \quad a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n}=n \cdot 3^n\end{aligned}$
$\begin{aligned}\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+a_3 x^3 & +\ldots \\ & +a_{2 \pi} x^{2 n}\end{aligned}$
On differentiating both sides, we get
$\begin{array}{r}n\left(1+x+x^2\right)^{n-1}(1+2 x)=a_1+2 a_2 x+3 a_3 x \\ +\ldots \ldots+2 n a_{2 n} x^{2 n-1}\end{array}$
Now, putting $x=1$, we get
$\begin{aligned} & n(3)^{n-1} \cdot(3)=a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n} \\ & \Rightarrow \quad a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n}=n \cdot 3^n\end{aligned}$
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