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

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

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sum from n=1 to infinity of 6/(2^n)

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

6

Solution steps

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

Apply the constant multiplication rule:

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

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

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

= 6 (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}})

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

= 6 (2 - \frac{1}{2 ^{0}})

Simplify

= 6

\frac{\partial}{\partial x} (y (1 - x))

e^{- 3 s} \cdot \frac{5}{4 s ^{2} + 16}

4 x - x^{2} (1.3)

\frac{d}{d x} (ln (e^{x} + e^{- x}))

x \to 4 + lim (\frac{2}{( x - 4 ) ^{4}})

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