What is the sum from n=3 to infinity of 4(1/2)^{n+2} ?

The sum from n=3 to infinity of 4(1/2)^{n+2} is 1/4

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sum from n=3 to infinity of 4(1/2)^{n+2}

n = 3 \sum \infty 4 (\frac{1}{2})^{n + 2}

\frac{1}{4}

Solution steps

n = 3 \sum \infty 4 (\frac{1}{2})^{n + 2}

Apply the constant multiplication rule:

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

= 4 \cdot n = 3 \sum \infty (\frac{1}{2})^{n + 2}

Simplify

(\frac{1}{2})^{n + 2} : (\frac{1}{2})^{n} (\frac{1}{2})^{2}

= 4 \cdot n = 3 \sum \infty (\frac{1}{2})^{n} (\frac{1}{2})^{2}

Apply the constant multiplication rule:

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

= 4 (\frac{1}{2})^{2} \cdot n = 3 \sum \infty (\frac{1}{2})^{n}

(\frac{1}{2})^{2} = \frac{1}{4}

= 4 \cdot \frac{1}{4} \cdot n = 3 \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} - (\frac{1}{2})^{1} - (\frac{1}{2})^{2})

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

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

Simplify

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

= \frac{1}{4}

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

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

n = 1 \sum \infty n e^{- 4 n^{2}}

n = 1 \sum \infty \frac{n !}{5 + n}

n = 5 \sum \infty \frac{1}{ln ( n )}

What is the sum from n=3 to infinity of 4(1/2)^{n+2} ?
The sum from n=3 to infinity of 4(1/2)^{n+2} is 1/4