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

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

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \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}

= 3 (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}

= 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}

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