\(\begin{aligned}
& \sum_{r=1}^n r^2 C_r=1^2 C_1+2^2 C_2+3^2 C_3+\ldots \ldots+n^2 C_n \\
& \because(1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots . .+C_n x^n
\end{aligned}\)
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.