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

The sum from n=1 to infinity of 7/(3^n) is 7/2

Popular >

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

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

\frac{7}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

n = 0 \sum \infty \frac{1}{3 ^{n}} = \frac{3}{2}

= 7 (\frac{3}{2} - \frac{1}{3 ^{0}})

Simplify

7 (\frac{3}{2} - \frac{1}{3 ^{0}}) : \frac{7}{2}

= \frac{7}{2}

\frac{d}{d x} (x^{2} + \frac{16}{x})

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

\frac{d}{d x} ((x + 1)^{s i n (x)})

2 \frac{d y}{d x} = y - e^{4 x}

f (x) = 6 x (6.6)

What is the sum from n=1 to infinity of 7/(3^n) ?
The sum from n=1 to infinity of 7/(3^n) is 7/2