What is the sum from n=2 to infinity of 5/(4^n) ?

The sum from n=2 to infinity of 5/(4^n) is 5/12

sum from n=2 to infinity of 5/(4^n)

n = 2 \sum \infty \frac{5}{4 ^{n}}

\frac{5}{12}

Decimal

0.41666 \dots

Solution steps

n = 2 \sum \infty \frac{5}{4 ^{n}}

Apply the constant multiplication rule:

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

= 5 \cdot n = 2 \sum \infty \frac{1}{4 ^{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}{4 ^{n}} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}})

n = 0 \sum \infty \frac{1}{4 ^{n}} = \frac{4}{3}

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

Simplify

5 (\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}}) : \frac{5}{12}

= \frac{5}{12}

\int_{4}^{10} 6 d x

\int k e^{- k^{2}}

\frac{d}{d x} (10 + (\frac{50 x}{2 x ^{2} + 9}))

x \to 0 - lim (\frac{6}{x} - \frac{6}{∣ x ∣})

x \cdot y^{^{'}} + y = x^{4} \cdot y^{3}

What is the sum from n=2 to infinity of 5/(4^n) ?
The sum from n=2 to infinity of 5/(4^n) is 5/12