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

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

Popular >

sum from k=1 to infinity of 3/(4^k)

k = 1 \sum \infty \frac{3}{4 ^{k}}

1

Solution steps

k = 1 \sum \infty \frac{3}{4 ^{k}}

Apply the constant multiplication rule:

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

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

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

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

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

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

Simplify

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

= 1

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

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

n = 0 \sum \infty n^{- \frac{2}{3}}

n = 1 \sum \infty 3 (- \frac{1}{8})^{3 n}

k = 3 \sum \infty \frac{4}{k} - \frac{4}{k + 1}

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