What is the sum from n=2 to infinity of 5/(7^n) ?

The sum from n=2 to infinity of 5/(7^n) is 5/42

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sum from n=2 to infinity of 5/(7^n)

n = 2 \sum \infty \frac{5}{7 ^{n}}

\frac{5}{42}

Solution steps

n = 2 \sum \infty \frac{5}{7 ^{n}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 2 \sum \infty \frac{1}{7 ^{n}}

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

= 5 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

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

Simplify

5 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{5}{42}

= \frac{5}{42}

\frac{\partial}{\partial y} (4 x^{2} y^{3})

\frac{\partial}{\partial x} (ln (x))

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{e ^{x}})

\frac{d y}{d x} = x^{4} y^{- 5}

y = 3 x e^{- k x}

What is the sum from n=2 to infinity of 5/(7^n) ?
The sum from n=2 to infinity of 5/(7^n) is 5/42