What is the sum from n=1 to infinity of (2022)/(9^n) ?

The sum from n=1 to infinity of (2022)/(9^n) is 1011/4

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sum from n=1 to infinity of (2022)/(9^n)

n = 1 \sum \infty \frac{2022}{9 ^{n}}

\frac{1011}{4}

Solution steps

n = 1 \sum \infty \frac{2022}{9 ^{n}}

Apply the constant multiplication rule:

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

= 2022 \cdot n = 1 \sum \infty \frac{1}{9 ^{n}}

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

= 2022 (n = 0 \sum \infty \frac{1}{9 ^{n}} - \frac{1}{9 ^{0}})

n = 0 \sum \infty \frac{1}{9 ^{n}} = \frac{9}{8}

= 2022 (\frac{9}{8} - \frac{1}{9 ^{0}})

Simplify

2022 (\frac{9}{8} - \frac{1}{9 ^{0}}) : \frac{1011}{4}

= \frac{1011}{4}

n = 1 \sum \infty n^{- 4}

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

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

k = 2 \sum \infty (\frac{1}{4})^{2 k}

n = 1 \sum \infty \frac{9 n}{6 n + 1}

What is the sum from n=1 to infinity of (2022)/(9^n) ?
The sum from n=1 to infinity of (2022)/(9^n) is 1011/4