What is the sum from m=2 to infinity of 3/(4^m) ?

The sum from m=2 to infinity of 3/(4^m) is 1/4

Popular >

sum from m=2 to infinity of 3/(4^m)

m = 2 \sum \infty \frac{3}{4 ^{m}}

\frac{1}{4}

Solution steps

m = 2 \sum \infty \frac{3}{4 ^{m}}

Apply the constant multiplication rule:

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

= 3 \cdot m = 2 \sum \infty \frac{1}{4 ^{m}}

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

= 3 (m = 0 \sum \infty \frac{1}{4 ^{m}} - \frac{1}{4 ^{0}} - \frac{1}{4 ^{1}})

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

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

Simplify

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

= \frac{1}{4}

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

n = 1 \sum \infty cos (\frac{3}{n ^{2}})

k = 4 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

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

n = 1 \sum \infty \frac{e ^{4 n}}{n !}

What is the sum from m=2 to infinity of 3/(4^m) ?
The sum from m=2 to infinity of 3/(4^m) is 1/4