What is the sum from n=1 to infinity of 3(2/3)^n ?

The sum from n=1 to infinity of 3(2/3)^n is 6

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sum from n=1 to infinity of 3(2/3)^n

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

6

Solution steps

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

Apply the constant multiplication rule:

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

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

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

= 3 (n = 0 \sum \infty (\frac{2}{3})^{n} - (\frac{2}{3})^{0})

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

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

Simplify

= 6

n = 1 \sum \infty 60 (0.1)^{n}

n = 0 \sum \infty (\frac{5}{9})^{n}

n = 1 \sum \infty \frac{n !}{9 4 ^{n}}

n = 0 \sum \infty 3 + (- 1)^{n}

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

What is the sum from n=1 to infinity of 3(2/3)^n ?
The sum from n=1 to infinity of 3(2/3)^n is 6