What is the sum from n=0 to infinity of 4/(2^{2n-2)} ?

The sum from n=0 to infinity of 4/(2^{2n-2)} is 64/3

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sum from n=0 to infinity of 4/(2^{2n-2)}

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

\frac{64}{3}

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

2^{2 n - 2} : 2^{2 n} \cdot 2^{- 2}

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

Apply the constant multiplication rule:

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

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

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

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

2^{2 n} = 4^{n}

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

Apply Series Geometric Test:

\frac{4}{3}

= 4 \cdot 4 \cdot \frac{4}{3}

Simplify

4 \cdot 4 \cdot \frac{4}{3} : \frac{64}{3}

= \frac{64}{3}

k = 2 \sum \infty k e^{- 3 k^{2}}

n = 0 \sum \infty \frac{1}{( 2 + n ) ^{2}}

n = 0 \sum \infty n^{2} e^{n}

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

n = 1 \sum \infty \frac{2}{3 n + 3}

What is the sum from n=0 to infinity of 4/(2^{2n-2)} ?
The sum from n=0 to infinity of 4/(2^{2n-2)} is 64/3