What is the sum from n=1 to infinity of (70)/(7^n) ?

The sum from n=1 to infinity of (70)/(7^n) is 35/3

Popular >

sum from n=1 to infinity of (70)/(7^n)

n = 1 \sum \infty \frac{70}{7 ^{n}}

\frac{35}{3}

Solution steps

n = 1 \sum \infty \frac{70}{7 ^{n}}

Apply the constant multiplication rule:

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

= 70 \cdot n = 1 \sum \infty \frac{1}{7 ^{n}}

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

= 70 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}})

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

= 70 (\frac{7}{6} - \frac{1}{7 ^{0}})

Simplify

70 (\frac{7}{6} - \frac{1}{7 ^{0}}) : \frac{35}{3}

= \frac{35}{3}

\int sec^{4} (\frac{x}{2}) d x

\int_{0}^{3} 30 (25 - t^{2}) d t

x \to \infty lim (∣ x + 2 ∣)

\int_{0}^{\infty} \frac{1}{2 x + 5} d x

\int_{0}^{1} \frac{1}{2 t ^{2} - 2 t + 1} d t

What is the sum from n=1 to infinity of (70)/(7^n) ?
The sum from n=1 to infinity of (70)/(7^n) is 35/3