29.Calculate $$ \sum_{k=1}^{\infty} 8\left(\frac{1}{2}\right)^{k+2} $$

Question

UpStudy AI · Accepted Answer

To calculate the infinite series:

$$
\sum_{k=1}^{\infty} 8\left(\frac{1}{2}\right)^{k+2}
$$

Let's break it down step by step.

1. **Simplify the Expression Inside the Summation:**

$$
   8\left(\frac{1}{2}\right)^{k+2} = 8 \times \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^k = 8 \times \frac{1}{4} \times \left(\frac{1}{2}\right)^k = 2 \times \left(\frac{1}{2}\right)^k
   $$

So, the series becomes:

$$
   \sum_{k=1}^{\infty} 2\left(\frac{1}{2}\right)^k
   $$

2. **Factor Out the Constant:**

$$
   2 \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^k
   $$

3. **Evaluate the Geometric Series:**

The sum of an infinite geometric series $$\sum_{k=1}^{\infty} r^k$$ is given by $$\frac{r}{1 - r}$$ for $$|r| < 1$$.

Here, $$r = \frac{1}{2}$$:

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

4. **Multiply by the Factored Constant:**

$$
   2 \times 1 = 2
   $$

**Final Answer:**

$$
\boxed{2}
$$
The sum of the series is 2.

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