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

The sum from n=1 to infinity of 4(3/4)^n is 12

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

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

12

Solution steps

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

Apply the constant multiplication rule:

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

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

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

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

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

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

Simplify

= 12

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What is the sum from n=1 to infinity of 4(3/4)^n ?
The sum from n=1 to infinity of 4(3/4)^n is 12