If ' $n$ ' is a positive integer, then $\sum_{r=1}^n r^2 \cdot C_r=(\ldots \ldots \ldots) 2^{n-2}$

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Aicharya · Accepted Answer

$\begin{aligned} & \sum_{r=1}^n r^2 C_r=1^2 C_1+2^2 C_2+3^2 C_3+\ldots \ldots+n^2 C_n \ & \because(1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots . .+C_n x^n \end{aligned}$ On differentiating both sides w.r.t. $x$, we get $n(1+x)^{n-1}=1 \cdot C_1+2 C_2 x+3 C_3 x^2+\ldots \ldots .+n C_n x^{n-1}$ Now, on multiplying by $x$ both sides, we get $n x(1+x)^{n-1}=1 . C_1 x+2 C_2 x^2+3 C_3 x^3+\ldots \ldots+n C_n x^n$ Now again on differentiating both sides w.r.t $x$, we get $\begin{aligned} & n\left[(1+x)^{n-1}+(n-1) x(1+x)^{n-2}\right] \ & \quad=1^2 C_1+2^2 C_2 x+3^2 C_3 x^2+\ldots \ldots+n^2 C_n x^{n-1} \end{aligned}$ Put $x=1$, we get $\begin{aligned} & 1^2 \cdot C_1+2^2 C_2+3^2 C_3+\ldots \ldots+n^2 C_n \ &= n\left[2^{n-1}+(n-1) 2^{n-2}\right] \ &= n 2^{n-2}[2+n-1]=n(n+1) 2^{n-2} \end{aligned}$ Hence, option (c) is correct.

If ' \(n\) ' is a positive integer, then \(\sum_{r=1}^n r^2 \cdot C_r=(\ldots \ldots \ldots) 2^{n-2}\)

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