What is the sum from n=1 to infinity of 53*(1/2)^n ?

The sum from n=1 to infinity of 53*(1/2)^n is 53

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sum from n=1 to infinity of 53*(1/2)^n

n = 1 \sum \infty 53 \cdot (\frac{1}{2})^{n}

53

Solution steps

n = 1 \sum \infty 53 (\frac{1}{2})^{n}

Apply the constant multiplication rule:

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

= 53 \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}

= 53 (n = 0 \sum \infty (\frac{1}{2})^{n} - (\frac{1}{2})^{0})

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

= 53 (2 - (\frac{1}{2})^{0})

Simplify

= 53

n = 0 \sum \infty \frac{1}{ln ( 6 n )}

n = 1 \sum \infty \frac{1}{12 n + 1}

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

n = 1 \sum \infty (- \frac{6}{25})^{n}

n = 0 \sum \infty \frac{4}{3 n + 2}

What is the sum from n=1 to infinity of 53*(1/2)^n ?
The sum from n=1 to infinity of 53*(1/2)^n is 53