Sequences & Series Question - None

Question

Let k be natural number. Defined $ S_k $ as the sum of the infinite geometric series with first term $ (k^2 - 1) $ and common ratio $ \dfrac{1}{k} $, that is $ S_k = \dfrac{k^2-1}{k^0} + \dfrac{k^2-1}{k^1} + \dfrac{k^2-1}{k^2} + \cdots $. The value of $ \displaystyle\sum_{k=1}^{\infty} \dfrac{S_k}{2^{k-1}} $, is:

Accepted Answer

Step 1: Find $S_k$ using the geometric series formula.

For an infinite geometric series with first term $a$ and common ratio $r$ where $|r| < 1$: $S = \frac{a}{1-r}$

Here, $a = k^2-1$ and $r = \frac{1}{k}$, so:

$$S_k = \frac{k^2-1}{1-\frac{1}{k}} = \frac{k^2-1}{\frac{k-1}{k}} = \frac{k(k^2-1)}{k-1} = \frac{k(k-1)(k+1)}{k-1} = k(k+1)$$

Step 2: Substitute into the required sum.

$$\sum_{k=1}^{\infty} \frac{S_k}{2^{k-1}} = \sum_{k=1}^{\infty} \frac{k(k+1)}{2^{k-1}}$$

Step 3: Evaluate the sum using standard techniques.

$$\sum_{k=1}^{\infty} \frac{k(k+1)}{2^{k-1}} = \sum_{k=1}^{\infty} \frac{k^2 + k}{2^{k-1}} = \sum_{k=1}^{\infty} \frac{k^2}{2^{k-1}} + \sum_{k=1}^{\infty} \frac{k}{2^{k-1}}$$

Using standard results: $\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = 4$ and $\sum_{k=1}^{\infty} \frac{k^2}{2^{k-1}} = 12$

$$= 12 + 4 = 16$$

∴ Answer: A (16)

Question:

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