What is the sum from n=1 to infinity of 3(1/2)^n ?

The sum from n=1 to infinity of 3(1/2)^n is 3

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sum from n=1 to infinity of 3(1/2)^n

n = 1 \sum \infty 3 (\frac{1}{2})^{n}

3

Solution steps

n = 1 \sum \infty 3 (\frac{1}{2})^{n}

Apply the constant multiplication rule:

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

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

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

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

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

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

Simplify

= 3

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

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

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

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

k = 1 \sum \infty k e^{- k}

What is the sum from n=1 to infinity of 3(1/2)^n ?
The sum from n=1 to infinity of 3(1/2)^n is 3