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

The sum from n=1 to infinity of 4/(2^{n+2)} is 1

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sum from n=1 to infinity of 4/(2^{n+2)}

n = 1 \sum \infty \frac{4}{2 ^{n + 2}}

1

Solution steps

n = 1 \sum \infty \frac{4}{2 ^{n + 2}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{2 ^{n + 2}}

Simplify

2^{n + 2} : 2^{n} \cdot 2^{2}

= 4 \cdot n = 1 \sum \infty \frac{1}{2 ^{n} \cdot 2 ^{2}}

Apply the constant multiplication rule:

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

= 4 \cdot \frac{1}{2 ^{2}} \cdot n = 1 \sum \infty \frac{1}{2 ^{n}}

Simplify

= 4 \cdot \frac{1}{4} \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}

= 4 \cdot \frac{1}{4} (n = 0 \sum \infty \frac{1}{2 ^{n}} - \frac{1}{2 ^{0}})

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

= 4 \cdot \frac{1}{4} (2 - \frac{1}{2 ^{0}})

Simplify

4 \cdot \frac{1}{4} (2 - \frac{1}{2 ^{0}}) : 1

= 1

n = 1 \sum \infty \frac{n !}{11 2 ^{n}}

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{2}}}

k = 1 \sum \infty k^{2} e^{- k}

n = 1 \sum \infty n! (- e)^{- 4 n}

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

What is the sum from n=1 to infinity of 4/(2^{n+2)} ?
The sum from n=1 to infinity of 4/(2^{n+2)} is 1