What is the sum from n=1 to infinity of (10)/(3^n) ?

The sum from n=1 to infinity of (10)/(3^n) is 5

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sum from n=1 to infinity of (10)/(3^n)

n = 1 \sum \infty \frac{10}{3 ^{n}}

5

Solution steps

n = 1 \sum \infty \frac{10}{3 ^{n}}

Apply the constant multiplication rule:

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

= 10 \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}

= 10 (n = 0 \sum \infty \frac{1}{3 ^{n}} - \frac{1}{3 ^{0}})

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

= 10 (\frac{3}{2} - \frac{1}{3 ^{0}})

Simplify

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

= 5

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

\int_{0}^{1} \frac{1}{( x + 1 ) ( x - 2 )} d x

\int \frac{x ^{2} - a ^{2}}{x} d x

\frac{\partial}{\partial y} (cos (x^{2}) sin (yz))

\frac{d y}{d x} = \frac{y ^{2} + 4 x x ^{2} + y ^{2}}{x y}

What is the sum from n=1 to infinity of (10)/(3^n) ?
The sum from n=1 to infinity of (10)/(3^n) is 5