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

1 Answer
Wataru
Apr 17, 2017

#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.

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