What is the sum from n=1 to infinity of (40)/(4^n) ?

The sum from n=1 to infinity of (40)/(4^n) is 40/3

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sum from n=1 to infinity of (40)/(4^n)

n = 1 \sum \infty \frac{40}{4 ^{n}}

\frac{40}{3}

Solution steps

n = 1 \sum \infty \frac{40}{4 ^{n}}

Apply the constant multiplication rule:

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

= 40 \cdot n = 1 \sum \infty \frac{1}{4 ^{n}}

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

= 40 (n = 0 \sum \infty \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}})

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= 40 (\frac{4}{3} - \frac{1}{4 ^{0}})

Simplify

40 (\frac{4}{3} - \frac{1}{4 ^{0}}) : \frac{40}{3}

= \frac{40}{3}

\frac{\partial}{\partial x} (x^{2} + 8 y^{2} + 9 z^{2} - 7)

\frac{d}{d x} (f (x) (5 x^{2} + 3))

\int x^{3} x^{4} + 10 d x

4 y^{^{''}} + 4 y^{^{'}} + 9 y = 0

\int_{2}^{3} \frac{1}{x ^{2} - 1} d x

What is the sum from n=1 to infinity of (40)/(4^n) ?
The sum from n=1 to infinity of (40)/(4^n) is 40/3