What is the sum from n=0 to infinity of 6(3/4)^n ?

The sum from n=0 to infinity of 6(3/4)^n is 24

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sum from n=0 to infinity of 6(3/4)^n

n = 0 \sum \infty 6 (\frac{3}{4})^{n}

24

Solution steps

n = 0 \sum \infty 6 (\frac{3}{4})^{n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 0 \sum \infty (\frac{3}{4})^{n}

Apply Series Geometric Test:

4

= 6 \cdot 4

Simplify

= 24

\int (ln (11 x))^{2} d x

3 x + 7

- 2.25 cos (1.5 x)

\int_{0}^{33} (x - 1)^{- \frac{1}{5}} d x

y = (x^{2} + 4) (x^{3} - 4 x), (- 2, 0)

What is the sum from n=0 to infinity of 6(3/4)^n ?
The sum from n=0 to infinity of 6(3/4)^n is 24