What is the sum from n=1 to infinity of (2^n)/(10^n) ?

The sum from n=1 to infinity of (2^n)/(10^n) is 1/4

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sum from n=1 to infinity of (2^n)/(10^n)

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

\frac{1}{4}

Solution steps

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{2 ^{n}}{1 0 ^{n}} = (\frac{2}{10})^{n}

= n = 1 \sum \infty (\frac{2}{10})^{n}

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

= n = 0 \sum \infty (\frac{2}{10})^{n} - (\frac{2}{10})^{0}

n = 0 \sum \infty (\frac{2}{10})^{n} = \frac{5}{4}

= \frac{5}{4} - (\frac{2}{10})^{0}

Simplify

\frac{5}{4} - (\frac{2}{10})^{0} : \frac{1}{4}

= \frac{1}{4}

\frac{\partial ^{2}}{\partial y ^{2}} (x y^{10} + x^{2} + y^{4})

f (x) = x^{- \frac{1}{2}} + x^{\frac{1}{2}}

\frac{1}{s ^{3} + s ^{2} + s}

y = x^{3} - x

f (x) = cos (4 t + 12)

What is the sum from n=1 to infinity of (2^n)/(10^n) ?
The sum from n=1 to infinity of (2^n)/(10^n) is 1/4