What is the sum from n=1 to infinity of 9/(7^n) ?

The sum from n=1 to infinity of 9/(7^n) is 3/2

Popular >

sum from n=1 to infinity of 9/(7^n)

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

\frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{7 ^{n}}

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

= 9 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}})

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

= 9 (\frac{7}{6} - \frac{1}{7 ^{0}})

Simplify

9 (\frac{7}{6} - \frac{1}{7 ^{0}}) : \frac{3}{2}

= \frac{3}{2}

n = 5 \sum \infty \frac{4}{n ^{2} - n}

n = 2 \sum \infty \frac{3}{n ^{3}}

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

i = 1 \sum \infty (\frac{1}{2})^{i}

n = 0 \sum \infty \frac{1}{8 - n}

What is the sum from n=1 to infinity of 9/(7^n) ?
The sum from n=1 to infinity of 9/(7^n) is 3/2