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

The sum from n=1 to infinity of 7/(5^n) is 7/4

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

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

\frac{7}{4}

Solution steps

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

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{5 ^{n}}

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

= 7 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}})

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= 7 (\frac{5}{4} - \frac{1}{5 ^{0}})

Simplify

7 (\frac{5}{4} - \frac{1}{5 ^{0}}) : \frac{7}{4}

= \frac{7}{4}

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

n = 1 \sum \infty \frac{1}{4 n ^{2} - 25}

m = 2 \sum \infty \frac{2}{7 ^{m}}

n = 0 \sum \infty \frac{- 2}{n + 1}

n = 1 \sum \infty (- \frac{7}{9})^{n - 1}

What is the sum from n=1 to infinity of 7/(5^n) ?
The sum from n=1 to infinity of 7/(5^n) is 7/4