Question

How do you evaluate `\sum _ { n = 1} ^ { \infty } 8( \frac { 1} { 2} ) ^ { k - 1}`?

Answer 1

`16`

Explanation:

Recall: (Geometric Series Sum Formula)

If `|r| < 1`, then `sum_(n=1)^(infty)ar^(n-1)=a/(1-r)`

By the formula above with `a=8` and `r=1/2`,

`sum_(n=1)^(infty)8(1/2)^(n-1)=8/(1-1/2)=16`

I hope that this was clear.