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

The sum from n=1 to infinity of 5/(3^n) is 5/2

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

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

\frac{5}{2}

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \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}

= 5 (n = 0 \sum \infty \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}})

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

= 5 (\frac{3}{2} - \frac{1}{3 ^{0}})

Simplify

5 (\frac{3}{2} - \frac{1}{3 ^{0}}) : \frac{5}{2}

= \frac{5}{2}

\frac{( e ^{x} + 1 ) d y}{d x} = y - y e^{x}

y = \frac{400}{x} + 7 + \frac{1}{5} x

x \to 4 lim (2 x - 6)

\int_{5}^{\infty} e^{- \frac{x}{2}} d x

\frac{d}{d x} (\frac{2}{3} x^{3} - 4 x^{2} + 10 x + 6)

What is the sum from n=1 to infinity of 5/(3^n) ?
The sum from n=1 to infinity of 5/(3^n) is 5/2