What is the sum from n=0 to infinity of 6/(2^n) ?

The sum from n=0 to infinity of 6/(2^n) is 12

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sum from n=0 to infinity of 6/(2^n)

n = 0 \sum \infty \frac{6}{2 ^{n}}

12

Solution steps

n = 0 \sum \infty \frac{6}{2 ^{n}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 0 \sum \infty \frac{1}{2 ^{n}}

Apply Series Geometric Test:

2

= 6 \cdot 2

Simplify

= 12

f (x) = \frac{- 6}{x}

\frac{\partial}{\partial x} (- 8 x y - 2 x^{4} - 2 y^{4})

x \to 0 lim (\frac{6 ^{x} - 5 ^{x}}{x})

f (x) = \frac{x - 3}{x + 3}

\frac{\partial}{\partial y} (\frac{x}{5 + y})

What is the sum from n=0 to infinity of 6/(2^n) ?
The sum from n=0 to infinity of 6/(2^n) is 12