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

The sum from n=2 to infinity of 7/(3^n) is 7/6

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sum from n=2 to infinity of 7/(3^n)

n = 2 \sum \infty \frac{7}{3 ^{n}}

\frac{7}{6}

Solution steps

n = 2 \sum \infty \frac{7}{3 ^{n}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 2 \sum \infty \frac{1}{3 ^{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}{3 ^{n}} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= 7 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

Simplify

7 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}}) : \frac{7}{6}

= \frac{7}{6}

n = 2 \sum \infty 2^{- 4 n}

n = 0 \sum \infty \frac{n}{6 + n ^{4}}

n = 1 \sum \infty n^{5} e^{n^{6}}

n = 5 \sum \infty n^{- s i n (3)}

n = 1 \sum \infty \frac{5}{n ^{10}}

What is the sum from n=2 to infinity of 7/(3^n) ?
The sum from n=2 to infinity of 7/(3^n) is 7/6