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

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

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sum from n=1 to infinity of 7/(2^n)

n = 1 \sum \infty \frac{7}{2 ^{n}}

7

Solution steps

n = 1 \sum \infty \frac{7}{2 ^{n}}

Apply the constant multiplication rule:

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

= 7 \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}

= 7 (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}})

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

= 7 (2 - \frac{1}{2 ^{0}})

Simplify

= 7

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