What is the sum from k=2 to infinity of 3/(2^k) ?

The sum from k=2 to infinity of 3/(2^k) is 3/2

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sum from k=2 to infinity of 3/(2^k)

k = 2 \sum \infty \frac{3}{2 ^{k}}

\frac{3}{2}

Solution steps

k = 2 \sum \infty \frac{3}{2 ^{k}}

Apply the constant multiplication rule:

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

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

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

= 3 (k = 0 \sum \infty \frac{1}{2 ^{k}} - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

k = 0 \sum \infty \frac{1}{2 ^{k}} = 2

= 3 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

Simplify

3 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}}) : \frac{3}{2}

= \frac{3}{2}

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What is the sum from k=2 to infinity of 3/(2^k) ?
The sum from k=2 to infinity of 3/(2^k) is 3/2