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

The sum from n=3 to infinity of 2/(5^n) is 1/50

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

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

\frac{1}{50}

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot n = 3 \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}

= 2 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} - \frac{1}{5 ^{2}})

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

= 2 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} - \frac{1}{5 ^{2}})

Simplify

2 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} - \frac{1}{5 ^{2}}) : \frac{1}{50}

= \frac{1}{50}

x \to 0 + lim (x + ∣ x ∣)

\frac{d}{d x} ((3 x^{2} + 4)^{2} (4 x^{2} - 8)^{2})

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (1) = 0, y^{^{'}} (1) = 1

(x^{4} + 3 x^{2} - 2)^{5}

(x + 1) y^{^{'}} + (x + 2) y = 8 x e^{- x}

What is the sum from n=3 to infinity of 2/(5^n) ?
The sum from n=3 to infinity of 2/(5^n) is 1/50