What is the sum from n=1 to infinity of 8(1/4)^n ?

The sum from n=1 to infinity of 8(1/4)^n is 8/3

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sum from n=1 to infinity of 8(1/4)^n

n = 1 \sum \infty 8 (\frac{1}{4})^{n}

\frac{8}{3}

Solution steps

n = 1 \sum \infty 8 (\frac{1}{4})^{n}

Apply the constant multiplication rule:

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

= 8 \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}

= 8 (n = 0 \sum \infty (\frac{1}{4})^{n} - (\frac{1}{4})^{0})

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

= 8 (\frac{4}{3} - (\frac{1}{4})^{0})

Simplify

8 (\frac{4}{3} - (\frac{1}{4})^{0}) : \frac{8}{3}

= \frac{8}{3}

\int c e^{c x} d x

\frac{\partial}{\partial x} (r cos (x) sin (y))

\frac{d y}{d t} = k y (1 - \frac{y}{4600})

f (x) = 8 sin (x), g (x) = 8 cos (2 x)

f (x) = \frac{1}{6} (2 x + 1)^{3}

What is the sum from n=1 to infinity of 8(1/4)^n ?
The sum from n=1 to infinity of 8(1/4)^n is 8/3