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

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

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

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

1

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \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})

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

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

Simplify

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

= 1

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

n = 1 \sum \infty \frac{4 n}{n ^{2} + 5}

n = 0 \sum \infty (\frac{5}{6})^{n - 1}

n = 1 \sum \infty (- \frac{1}{4})^{n - 1}

n = 0 \sum \infty \frac{6}{7 ^{n}}

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