What is the sum from n=0 to infinity of 6(3/4)^n ?

The sum from n=0 to infinity of 6(3/4)^n is 24

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sum from n=0 to infinity of 6(3/4)^n

n = 0 \sum \infty 6 (\frac{3}{4})^{n}

24

Solution steps

n = 0 \sum \infty 6 (\frac{3}{4})^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 6 \cdot n = 0 \sum \infty (\frac{3}{4})^{n}

Apply Series Geometric Test:

4

= 6 \cdot 4

Simplify

= 24

What is the sum from n=0 to infinity of 6(3/4)^n ?
The sum from n=0 to infinity of 6(3/4)^n is 24