What is the sum from n=2 to infinity of 7/(4^n) ?

The sum from n=2 to infinity of 7/(4^n) is 7/12

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sum from n=2 to infinity of 7/(4^n)

n = 2 \sum \infty \frac{7}{4 ^{n}}

\frac{7}{12}

Solution steps

n = 2 \sum \infty \frac{7}{4 ^{n}}

Apply the constant multiplication rule:

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

= 7 \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}

= 7 (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}

= 7 (\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}})

Simplify

7 (\frac{4}{3} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}}) : \frac{7}{12}

= \frac{7}{12}

x \to \infty lim (e^{x} x)

\int \frac{3}{x + 4} d x

x \to 4 lim ((2 x - 3)^{2})

\int_{0}^{2} \frac{1}{x ^{2} - 4} d x

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

What is the sum from n=2 to infinity of 7/(4^n) ?
The sum from n=2 to infinity of 7/(4^n) is 7/12