What is the sum from n=1 to infinity of 9(4/5)^n ?

The sum from n=1 to infinity of 9(4/5)^n is 36

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sum from n=1 to infinity of 9(4/5)^n

n = 1 \sum \infty 9 (\frac{4}{5})^{n}

36

Solution steps

n = 1 \sum \infty 9 (\frac{4}{5})^{n}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty (\frac{4}{5})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 9 (n = 0 \sum \infty (\frac{4}{5})^{n} - (\frac{4}{5})^{0})

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

= 9 (5 - (\frac{4}{5})^{0})

Simplify

= 36

n = 0 \sum \infty 2 (\frac{1}{4})^{n}

n = 4 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

n = 0 \sum \infty \frac{n}{11 n + 1}

m = 2 \sum \infty \frac{3}{4 ^{m}}

n = 1 \sum \infty \frac{n ^{2}}{5 + 2 n}

What is the sum from n=1 to infinity of 9(4/5)^n ?
The sum from n=1 to infinity of 9(4/5)^n is 36