What is the sum from n=2 to infinity of 3/(6^n) ?

The sum from n=2 to infinity of 3/(6^n) is 1/10

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sum from n=2 to infinity of 3/(6^n)

n = 2 \sum \infty \frac{3}{6 ^{n}}

\frac{1}{10}

Solution steps

n = 2 \sum \infty \frac{3}{6 ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 2 \sum \infty \frac{1}{6 ^{n}}

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

= 3 (n = 0 \sum \infty \frac{1}{6 ^{n}} - \frac{1}{6 ^{0}} - \frac{1}{6 ^{1}})

n = 0 \sum \infty \frac{1}{6 ^{n}} = \frac{6}{5}

= 3 (\frac{6}{5} - \frac{1}{6 ^{0}} - \frac{1}{6 ^{1}})

Simplify

3 (\frac{6}{5} - \frac{1}{6 ^{0}} - \frac{1}{6 ^{1}}) : \frac{1}{10}

= \frac{1}{10}

x \to 0 lim (e^{x} cos (5 x))

81 y^{^{''}} + 72 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

e^{\frac{2 t}{1}}

\frac{6 s ^{2}}{s ^{2} ( s ^{2} + 16 )}

\frac{\partial}{\partial x} (\frac{3 x - 5}{2 x + 4})

What is the sum from n=2 to infinity of 3/(6^n) ?
The sum from n=2 to infinity of 3/(6^n) is 1/10