What is the sum from n=2 to infinity of 7/(2^n) ?

The sum from n=2 to infinity of 7/(2^n) is 7/2

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sum from n=2 to infinity of 7/(2^n)

n = 2 \sum \infty \frac{7}{2 ^{n}}

\frac{7}{2}

Solution steps

n = 2 \sum \infty \frac{7}{2 ^{n}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 2 \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}

= 7 (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

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

= 7 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}})

Simplify

7 (2 - \frac{1}{2 ^{0}} - \frac{1}{2 ^{1}}) : \frac{7}{2}

= \frac{7}{2}

\int_{0}^{1} \frac{32}{x ^{2} - 64} d x

\frac{1}{s ^{2} \cdot ( s + 1 )}

f (x) = \frac{7}{2} x^{\frac{2}{7}}

x \to 8 lim (\frac{x ^{2} - 64}{x ^{2} - 8 x})

\frac{\partial}{\partial x} (2 x + 1)

What is the sum from n=2 to infinity of 7/(2^n) ?
The sum from n=2 to infinity of 7/(2^n) is 7/2