What is the sum from n=3 to infinity of 4/(7^n) ?

The sum from n=3 to infinity of 4/(7^n) is 2/147

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sum from n=3 to infinity of 4/(7^n)

n = 3 \sum \infty \frac{4}{7 ^{n}}

\frac{2}{147}

Solution steps

n = 3 \sum \infty \frac{4}{7 ^{n}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 3 \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}

= 4 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}} - \frac{1}{7 ^{2}})

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

= 4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}} - \frac{1}{7 ^{2}})

Simplify

4 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}} - \frac{1}{7 ^{2}}) : \frac{2}{147}

= \frac{2}{147}

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

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

k = 1 \sum \infty (- \frac{1}{3})^{k}

n = 1 \sum \infty (\frac{3}{5})^{n - 1}

n = 0 \sum \infty \frac{5}{5 - n}

What is the sum from n=3 to infinity of 4/(7^n) ?
The sum from n=3 to infinity of 4/(7^n) is 2/147