What is the sum from n=3 to infinity of 5(4/9)^n ?

The sum from n=3 to infinity of 5(4/9)^n is 64/81

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sum from n=3 to infinity of 5(4/9)^n

n = 3 \sum \infty 5 (\frac{4}{9})^{n}

\frac{64}{81}

Solution steps

n = 3 \sum \infty 5 (\frac{4}{9})^{n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 3 \sum \infty (\frac{4}{9})^{n}

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

= 5 (n = 0 \sum \infty (\frac{4}{9})^{n} - (\frac{4}{9})^{0} - (\frac{4}{9})^{1} - (\frac{4}{9})^{2})

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

= 5 (\frac{9}{5} - (\frac{4}{9})^{0} - (\frac{4}{9})^{1} - (\frac{4}{9})^{2})

Simplify

5 (\frac{9}{5} - (\frac{4}{9})^{0} - (\frac{4}{9})^{1} - (\frac{4}{9})^{2}) : \frac{64}{81}

= \frac{64}{81}

n = 1 \sum \infty \frac{5}{n ^{2} + 121}

n = 2 \sum \infty \frac{7 n}{ln ( n )}

n = 1 \sum \infty \frac{6 n ^{2}}{n !}

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

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

What is the sum from n=3 to infinity of 5(4/9)^n ?
The sum from n=3 to infinity of 5(4/9)^n is 64/81