What is the sum from n=2 to infinity of 1/(4^{2n)} ?

The sum from n=2 to infinity of 1/(4^{2n)} is 1/240

Popular >

sum from n=2 to infinity of 1/(4^{2n)}

n = 2 \sum \infty \frac{1}{4 ^{2 n}}

\frac{1}{240}

Solution steps

n = 2 \sum \infty \frac{1}{4 ^{2 n}}

4^{2 n} = 1 6^{n}

= n = 2 \sum \infty \frac{1}{1 6 ^{n}}

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

= n = 0 \sum \infty \frac{1}{1 6 ^{n}} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}}

n = 0 \sum \infty \frac{1}{1 6 ^{n}} = \frac{16}{15}

= \frac{16}{15} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}}

Simplify

\frac{16}{15} - \frac{1}{1 6 ^{0}} - \frac{1}{1 6 ^{1}} : \frac{1}{240}

= \frac{1}{240}

\int \frac{1}{4 x ^{2} + 12 x + 18} d x

\frac{\partial}{\partial x} (4 + x^{3} + y^{3} - 3 x y)

2 e^{- t} - \frac{3}{2} e^{- 2 t}

\frac{d}{d x} (sec (x) - 2 cos (x))

\int 8 sin (t) d t

What is the sum from n=2 to infinity of 1/(4^{2n)} ?
The sum from n=2 to infinity of 1/(4^{2n)} is 1/240