What is the sum from n=2 to infinity of (3^n)/(12^n) ?

The sum from n=2 to infinity of (3^n)/(12^n) is 1/12

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sum from n=2 to infinity of (3^n)/(12^n)

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

\frac{1}{12}

Solution steps

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

Simplify

\frac{3 ^{n}}{1 2 ^{n}} : \frac{1}{4 ^{n}}

= n = 2 \sum \infty \frac{1}{4 ^{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{1}{4 ^{n}} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}}

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

= \frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}}

Simplify

\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}} : \frac{1}{12}

= \frac{1}{12}

4 y^{^{''}} - 4 y^{^{'}} + y = 8 e^{\frac{x}{2}}

y = 8 e^{17 t}

4 csc (5 x)

f (t) = 4 t^{2}

\frac{\partial}{\partial x} (A e^{i (k x - wt)})

What is the sum from n=2 to infinity of (3^n)/(12^n) ?
The sum from n=2 to infinity of (3^n)/(12^n) is 1/12