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

The sum from n=1 to infinity of 5(3/7)^n is 15/4

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sum from n=1 to infinity of 5(3/7)^n

n = 1 \sum \infty 5 (\frac{3}{7})^{n}

\frac{15}{4}

Decimal

3.75

Solution steps

n = 1 \sum \infty 5 (\frac{3}{7})^{n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty (\frac{3}{7})^{n}

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

= 5 (n = 0 \sum \infty (\frac{3}{7})^{n} - (\frac{3}{7})^{0})

n = 0 \sum \infty (\frac{3}{7})^{n} = \frac{7}{4}

= 5 (\frac{7}{4} - (\frac{3}{7})^{0})

Simplify

5 (\frac{7}{4} - (\frac{3}{7})^{0}) : \frac{15}{4}

= \frac{15}{4}

What is the sum from n=1 to infinity of 5(3/7)^n ?
The sum from n=1 to infinity of 5(3/7)^n is 15/4