Question:

$\textcolor{green}{\textbf{Key Idea}}$ First compute the geometric sum: \[ S_k=\frac{k^2-1}{1-1/k}=k(k+1) \qquad (k\geq 2). \] But for \(k=1\), the first term itself is \(0\), so \[ S_1=0. \] Hence the required sum is \[ \sum_{k=2}^{\infty}\frac{k(k+1)}{2^{k-1}}. \] Use standard power-series sums for \(\sum k r^{k-1}\) and \(\sum k^2 r^{k-1}\). $\textcolor{blue}{\textbf{Solution}}$ For \(k\geq 2\), \(S_k\) is the sum of a geometric series with first term \(k^2-1\) and common ratio \(1/k\). Hence \[ S_k=\frac{k^2-1}{1-\frac{1}{k}}. \] Now \[ 1-\frac{1}{k}=\frac{k-1}{k}. \] Therefore \[ S_k =\frac{k^2-1}{(k-1)/k} =\frac{(k-1)(k+1)k}{k-1} =k(k+1). \] For \(k=1\), the first term is \[ k^2-1=0, \] so \[ S_1=0. \] Thus the required sum is \[ \sum_{k=2}^{\infty}\frac{k(k+1)}{2^{k-1}} \] Put \[ r=\frac12. \] Then \[ \sum_{k=1}^{\infty}k r^{k-1} =\frac{1}{(1-r)^2} \] and \[ \sum_{k=1}^{\infty}k^2 r^{k-1} =\frac{1+r}{(1-r)^3}. \] Hence \[ \sum_{k=1}^{\infty}k(k+1)r^{k-1} =\sum_{k=1}^{\infty}k^2r^{k-1} +\sum_{k=1}^{\infty}kr^{k-1}. \] At \(r=\frac12\), \[ \sum_{k=1}^{\infty}k^2r^{k-1} =\frac{1+\frac12}{\left(1-\frac12\right)^3} =12, \] and \[ \sum_{k=1}^{\infty}kr^{k-1} =\frac{1}{\left(1-\frac12\right)^2} =4. \] Therefore \[ \sum_{k=1}^{\infty}k(k+1)r^{k-1} =12+4=16. \] This includes the \(k=1\) contribution \[ 1(1+1)\left(\frac12\right)^0=2. \] But the actual value is \(S_1=0\), so we subtract this extra \(2\): \[ \sum_{k=1}^{\infty}\frac{S_k}{2^{k-1}} =16-2=14. \] \[ \boxed{14} \] $\textcolor{red}{\textbf{Key Trap}}$ The case \(k=1\) looks awkward because the common ratio is \(1\), but the first term is \[ k^2-1=0. \] So \(S_1=0\), and the compact formula \(S_k=k(k+1)\) contributes \(2\) at \(k=1\) only if used carelessly. Treat the series sum through the limiting formula for \(k\geq 2\), then note the actual first term is zero. The naive power-series sum from \(k=1\) gives \(16\), but the corrected answer is \[ 16-2=14. \]