What is the sum from n=1 to infinity of 5(7/8)^n ?

The sum from n=1 to infinity of 5(7/8)^n is 35

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sum from n=1 to infinity of 5(7/8)^n

n = 1 \sum \infty 5 (\frac{7}{8})^{n}

35

Solution steps

n = 1 \sum \infty 5 (\frac{7}{8})^{n}

Apply the constant multiplication rule:

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

= 5 \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}

= 5 (n = 0 \sum \infty (\frac{7}{8})^{n} - (\frac{7}{8})^{0})

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

= 5 (8 - (\frac{7}{8})^{0})

Simplify

= 35

\frac{d}{d x} (\frac{3 x}{x})

18 t^{2} - 11 t + 4

x^{2} y^{^{'}} + \frac{4}{3} x y = - 2 cos (4 x) y^{- \frac{1}{2}}

\int_{5}^{8} \frac{x}{x ^{2} - 2} d x

\frac{d}{d x} (\frac{a + x ^{2}}{a - x ^{2}})

What is the sum from n=1 to infinity of 5(7/8)^n ?
The sum from n=1 to infinity of 5(7/8)^n is 35