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

The sum from n=1 to infinity of 2(3/4)^n is 6

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

n = 1 \sum \infty 2 (\frac{3}{4})^{n}

6

Solution steps

n = 1 \sum \infty 2 (\frac{3}{4})^{n}

Apply the constant multiplication rule:

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

= 2 \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}

= 2 (n = 0 \sum \infty (\frac{3}{4})^{n} - (\frac{3}{4})^{0})

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

= 2 (4 - (\frac{3}{4})^{0})

Simplify

= 6

What is the sum from n=1 to infinity of 2(3/4)^n ?
The sum from n=1 to infinity of 2(3/4)^n is 6