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

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

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sum from n=1 to infinity of 9/(-2^n)

n = 1 \sum \infty \frac{9}{- 2 ^{n}}

- 9

Solution steps

n = 1 \sum \infty \frac{9}{- 2 ^{n}}

Apply the constant multiplication rule:

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

= \frac{9}{- 1} \cdot n = 1 \sum \infty \frac{1}{2 ^{n}}

Simplify

= (- 9) \cdot n = 1 \sum \infty \frac{1}{2 ^{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}{2 ^{n}} - \frac{1}{2 ^{0}})

n = 0 \sum \infty \frac{1}{2 ^{n}} = 2

= (- 9) (2 - \frac{1}{2 ^{0}})

Simplify

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

= - 9

n = 1 \sum \infty 45 (- 1)^{n}

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

n = 1 \sum \infty \frac{4 n}{3 n + 1}

n = 1 \sum \infty \frac{n + 1}{3 - 2 n}

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

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