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

The sum from n=2 to infinity of 3/(5^n) is 3/20

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

n = 2 \sum \infty \frac{3}{5 ^{n}}

\frac{3}{20}

Solution steps

n = 2 \sum \infty \frac{3}{5 ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 2 \sum \infty \frac{1}{5 ^{n}}

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

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

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= 3 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

Simplify

3 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}) : \frac{3}{20}

= \frac{3}{20}

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

\int (sin (x) + 2 cos (x)) d x

\frac{d}{d x} (x^{t})

\int_{0}^{4} (2 t + 4)^{- \frac{1}{2}} d t

y^{^{'}} + 2 y = y^{2}

What is the sum from n=2 to infinity of 3/(5^n) ?
The sum from n=2 to infinity of 3/(5^n) is 3/20