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

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

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sum from n=0 to infinity of 3/(2^{2n)}

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

4

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{2 ^{2 n}}

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

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

Apply Series Geometric Test:

\frac{4}{3}

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

Simplify

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

= 4

n = 1 \sum \infty (\frac{12}{8})^{n}

n = 1 \sum \infty (cos (19))^{n}

n = 1 \sum \infty \frac{1}{tan ( n )}

n = 2 \sum \infty \frac{1}{n \cdot ( n - 1 )}

n = 0 \sum \infty \frac{2 n ^{2}}{n !}

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