Question:

Step 1: Find the formula for $S_k$ using the infinite geometric series formula. For $k \geq 2$, we have an infinite geometric series with first term $a = k^2 - 1$ and common ratio $r = \frac{1}{k}$. Since $\left|\frac{1}{k}\right| < 1$ for $k \geq 2$, the series converges and we can use the formula: $$S_k = \frac{a}{1-r} = \frac{k^2-1}{1 - \frac{1}{k}} = \frac{k^2-1}{\frac{k-1}{k}} = \frac{k(k^2-1)}{k-1}$$ Step 2: Simplify the expression for $S_k$. Factor the numerator: $$S_k = \frac{k(k-1)(k+1)}{k-1} = k(k+1)$$ This holds for all $k \geq 2$. Step 3: Handle the special case $k=1$. When $k=1$, the first term is $1^2 - 1 = 0$, so: $$S_1 = 0$$ Step 4: Set up the sum we need to evaluate. $$\sum_{k=1}^{\infty} \frac{S_k}{2^{k-1}} = \frac{S_1}{2^0} + \sum_{k=2}^{\infty} \frac{S_k}{2^{k-1}} = 0 + \sum_{k=2}^{\infty} \frac{k(k+1)}{2^{k-1}}$$ Step 5: Use the standard formula for $\sum_{k=1}^{\infty} k(k+1)x^{k-1}$. We use the known result: $$\sum_{k=1}^{\infty} k(k+1)x^{k-1} = \frac{2}{(1-x)^3}$$ Step 6: Evaluate the sum at $x = \frac{1}{2}$. $$\sum_{k=1}^{\infty} k(k+1)\left(\frac{1}{2}\right)^{k-1} = \frac{2}{\left(1-\frac{1}{2}\right)^3} = \frac{2}{\left(\frac{1}{2}\right)^3} = \frac{2}{\frac{1}{8}} = 16$$ Step 7: Subtract the $k=1$ term to get the sum starting from $k=2$. The $k=1$ term in the sum $\sum_{k=1}^{\infty} k(k+1)\left(\frac{1}{2}\right)^{k-1}$ is: $$\frac{1 \cdot 2}{2^0} = 2$$ Therefore: $$\sum_{k=2}^{\infty} \frac{k(k+1)}{2^{k-1}} = 16 - 2 = 14$$ Step 8: State the final answer. $$\sum_{k=1}^{\infty} \frac{S_k}{2^{k-1}} = 14$$ The answer is **Option 4: 14**.