What is the sum from n=1 to infinity of 8(7/8)^n ?

The sum from n=1 to infinity of 8(7/8)^n is 56

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sum from n=1 to infinity of 8(7/8)^n

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

56

Solution steps

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

Apply the constant multiplication rule:

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

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

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

= 8 (n = 0 \sum \infty (\frac{7}{8})^{n} - (\frac{7}{8})^{0})

n = 0 \sum \infty (\frac{7}{8})^{n} = 8

= 8 (8 - (\frac{7}{8})^{0})

Simplify

= 56

\int 18 x^{2} (6 x^{3} + 5)^{\frac{1}{2}} d x

\int tan^{2} (x) (sec^{2} (x) - 1) d x

\frac{d}{d x} ((x^{3} + x)^{\frac{3}{2}})

\frac{d}{d x} (2^{c o s (x)})

n = 0 \sum \infty 2^{3 - 2 n}

What is the sum from n=1 to infinity of 8(7/8)^n ?
The sum from n=1 to infinity of 8(7/8)^n is 56