What is the sum from n=2 to infinity of 7/(6^n) ?

The sum from n=2 to infinity of 7/(6^n) is 7/30

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sum from n=2 to infinity of 7/(6^n)

n = 2 \sum \infty \frac{7}{6 ^{n}}

\frac{7}{30}

Solution steps

n = 2 \sum \infty \frac{7}{6 ^{n}}

Apply the constant multiplication rule:

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

= 7 \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}

= 7 (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}

= 7 (\frac{6}{5} - \frac{1}{6 ^{0}} - \frac{1}{6 ^{1}})

Simplify

7 (\frac{6}{5} - \frac{1}{6 ^{0}} - \frac{1}{6 ^{1}}) : \frac{7}{30}

= \frac{7}{30}

\int 23 arctan (x) d x

f (x) = cos (5 x^{2} + 5)

\frac{d y}{d x} = y x^{5} sin (x^{6})

y^{^{''}} + 5^{2} y = 0

\frac{d}{d x} (- \frac{1}{2})

What is the sum from n=2 to infinity of 7/(6^n) ?
The sum from n=2 to infinity of 7/(6^n) is 7/30