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

The sum from n=3 to infinity of 1/(4^n) is 1/48

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

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

\frac{1}{48}

Solution steps

n = 3 \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}

= n = 0 \sum \infty \frac{1}{4 ^{n}} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}}

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

= \frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}}

Simplify

\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} - \frac{1}{4 ^{2}} : \frac{1}{48}

= \frac{1}{48}

4 y^{^{''}} + 8 y^{^{'}} + 13 y = 8 sin (1.5 t)

\int \frac{1}{5 x ^{3}} d x

\frac{1}{5} x^{15}

\int \frac{x ^{2}}{( x ^{2} + 4 ) ^{2}} d x

(- 3, 4), (2, - 2)

What is the sum from n=3 to infinity of 1/(4^n) ?
The sum from n=3 to infinity of 1/(4^n) is 1/48