What is the sum from n=1 to infinity of 3(4/5)^n ?

The sum from n=1 to infinity of 3(4/5)^n is 12

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sum from n=1 to infinity of 3(4/5)^n

n = 1 \sum \infty 3 (\frac{4}{5})^{n}

12

Solution steps

n = 1 \sum \infty 3 (\frac{4}{5})^{n}

Apply the constant multiplication rule:

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

= 3 \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}

= 3 (n = 0 \sum \infty (\frac{4}{5})^{n} - (\frac{4}{5})^{0})

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

= 3 (5 - (\frac{4}{5})^{0})

Simplify

= 12

n = 0 \sum \infty (- \frac{1}{9})^{n}

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

n = 0 \sum \infty \frac{4}{1 + 3 n ^{3}}

n = 0 \sum \infty \frac{n ^{2}}{5 + 2 n}

n = 1 \sum \infty (sin (2))^{n}

What is the sum from n=1 to infinity of 3(4/5)^n ?
The sum from n=1 to infinity of 3(4/5)^n is 12