What is the sum from n=1 to infinity of-7(3/8)^n ?

The sum from n=1 to infinity of-7(3/8)^n is -21/5

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sum from n=1 to infinity of-7(3/8)^n

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

- \frac{21}{5}

Decimal

- 4.2

Solution steps

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

Apply the constant multiplication rule:

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

= - 7 \cdot n = 1 \sum \infty (\frac{3}{8})^{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{3}{8})^{n} - (\frac{3}{8})^{0})

n = 0 \sum \infty (\frac{3}{8})^{n} = \frac{8}{5}

= - 7 (\frac{8}{5} - (\frac{3}{8})^{0})

Simplify

- 7 (\frac{8}{5} - (\frac{3}{8})^{0}) : - \frac{21}{5}

= - \frac{21}{5}

What is the sum from n=1 to infinity of-7(3/8)^n ?
The sum from n=1 to infinity of-7(3/8)^n is -21/5