What is the sum from k=2 to infinity of 5(1/3)^k ?

The sum from k=2 to infinity of 5(1/3)^k is 5/6

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sum from k=2 to infinity of 5(1/3)^k

k = 2 \sum \infty 5 (\frac{1}{3})^{k}

\frac{5}{6}

Solution steps

k = 2 \sum \infty 5 (\frac{1}{3})^{k}

Apply the constant multiplication rule:

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

= 5 \cdot k = 2 \sum \infty (\frac{1}{3})^{k}

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

= 5 (k = 0 \sum \infty (\frac{1}{3})^{k} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1})

k = 0 \sum \infty (\frac{1}{3})^{k} = \frac{3}{2}

= 5 (\frac{3}{2} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1})

Simplify

5 (\frac{3}{2} - (\frac{1}{3})^{0} - (\frac{1}{3})^{1}) : \frac{5}{6}

= \frac{5}{6}

g = 1 \sum \infty g^{gx}

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

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

n = 2 \sum \infty (- 0.9)^{n}

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

What is the sum from k=2 to infinity of 5(1/3)^k ?
The sum from k=2 to infinity of 5(1/3)^k is 5/6