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 (\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{8}{n ( n + 2 )}

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

n = 1 \sum \infty \frac{1}{5 ^{n} - 1}

n = 0 \sum \infty (- \frac{4}{5})^{n}

n = 2 \sum \infty (- 0.35)^{n}

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