- #1

goohu

- 54

- 3

Hi, I'm trying to solve the sum of following infinite series:

\(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}+4}{{2}^{k}} = \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}}

+ \sum_{k=1}^{\infty} \frac{4}{{2}^{k}}\)

Using partial sum we can rewrite the first series: \(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}} = \sum_{k=1}^{n} \frac{{k}^{2}}{{2}^{k}} = \frac{n(n+1)(2n+1)}{6} * \left(\frac{1}{2}\right)^{\!{k}} \) , as n goes to \(\displaystyle \infty\)

The second series is a geometric one and can be rewritten as: \(\displaystyle \sum_{k=1}^{\infty} \frac{4}{{2}^{k}} = 4* \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{\!{k}} = 1/2 \)

It's unclear how to get the sum from the term \(\displaystyle \frac{n(n+1)(2n+1)}{6} \)

\(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}+4}{{2}^{k}} = \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}}

+ \sum_{k=1}^{\infty} \frac{4}{{2}^{k}}\)

Using partial sum we can rewrite the first series: \(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}} = \sum_{k=1}^{n} \frac{{k}^{2}}{{2}^{k}} = \frac{n(n+1)(2n+1)}{6} * \left(\frac{1}{2}\right)^{\!{k}} \) , as n goes to \(\displaystyle \infty\)

The second series is a geometric one and can be rewritten as: \(\displaystyle \sum_{k=1}^{\infty} \frac{4}{{2}^{k}} = 4* \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{\!{k}} = 1/2 \)

It's unclear how to get the sum from the term \(\displaystyle \frac{n(n+1)(2n+1)}{6} \)

Last edited: