What is the sum from n=0 to infinity of 7(4/5)^n ?

The sum from n=0 to infinity of 7(4/5)^n is 35

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sum from n=0 to infinity of 7(4/5)^n

n = 0 \sum \infty 7 (\frac{4}{5})^{n}

35

Solution steps

n = 0 \sum \infty 7 (\frac{4}{5})^{n}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty (\frac{4}{5})^{n}

Apply Series Geometric Test:

5

= 7 \cdot 5

Simplify

= 35

n = 1 \sum \infty \frac{n ^{2}}{8 ^{n}}

n = 1 \sum \infty \frac{9 ^{n}}{n + 1}

n = 0 \sum \infty (\frac{10}{11})^{n}

i = 0 \sum \infty \frac{1}{2 ^{i}}

n = 3 \sum \infty \frac{4}{7 ^{n}}

What is the sum from n=0 to infinity of 7(4/5)^n ?
The sum from n=0 to infinity of 7(4/5)^n is 35