What is the sum from k=1 to infinity of 5/(2^k) ?

The sum from k=1 to infinity of 5/(2^k) is 5

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sum from k=1 to infinity of 5/(2^k)

k = 1 \sum \infty \frac{5}{2 ^{k}}

5

Solution steps

k = 1 \sum \infty \frac{5}{2 ^{k}}

Apply the constant multiplication rule:

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

= 5 \cdot k = 1 \sum \infty \frac{1}{2 ^{k}}

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

= 5 (k = 0 \sum \infty \frac{1}{2 ^{k}} - \frac{1}{2 ^{0}})

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

= 5 (2 - \frac{1}{2 ^{0}})

Simplify

= 5

\frac{d y}{d x} = e^{x} + y

\int \frac{4}{x ^{5}} - 5 x^{4} d x

\frac{\partial}{\partial y} (ln (x^{6} y))

n = 0 \sum \infty (\frac{2}{3})^{n}

\frac{d}{d x} (- \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

What is the sum from k=1 to infinity of 5/(2^k) ?
The sum from k=1 to infinity of 5/(2^k) is 5