What is the sum from n=0 to infinity of 1/(1.02^n) ?

The sum from n=0 to infinity of 1/(1.02^n) is 51

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sum from n=0 to infinity of 1/(1.02^n)

n = 0 \sum \infty \frac{1}{1.0 2 ^{n}}

51

Solution steps

n = 0 \sum \infty \frac{1}{1.0 2 ^{n}}

Apply Series Geometric Test:

51

= 51

\frac{d}{d T} (\frac{7}{6} T + 17)

\frac{d y}{d x} = \frac{y x + y}{y x + x}

x d x - y d y = 0

\frac{\partial}{\partial z} (e^{y})

y = x^{4} - 9 x^{2}, y = x^{2} - 9

What is the sum from n=0 to infinity of 1/(1.02^n) ?
The sum from n=0 to infinity of 1/(1.02^n) is 51