1. ## Sum the series

Sorry, don't know how to use latex at the moment, but:

Summation from k=0 to infinity of (k+2)2^(-k)

$\sum (k+2)2^{-k}$

It says to sum the series.

I'm really unsure where to start in this problem. The only similarity I can notice is 1/(1-x) which equals the summation of x^k, but I don't think you can plug in -k's to change summations, can you?

Any ideas?

Edit-
Nvm, put it in LaTex too.

2. $\displaystyle \sum_{k\ge 0}\frac{k+2}{2^k} = \sum_{k\ge 0}\frac{k}{2^k}+\sum_{k\ge 0}\frac{2}{2^k}$. Putting $\displaystyle x = \frac{1}{2}$ in $\displaystyle \frac{2}{1-x} = 2\sum_{k\ge 0}x^k$, gives $\displaystyle \sum_{k\ge 0}\frac{2}{2^k} = 4$.
Putting $x = \frac{1}{2}$ in $\displaystyle \frac{x}{(x-1)^2} = \sum_{k \ge 0}kx^k$ gives $\displaystyle \sum_{k \ge 0}\frac{k}{2^k} = 2$. Therefore $\displaystyle \sum_{k\ge 0}\frac{k+2}{2^k} = 6.$

3. Originally Posted by ElTaco
Sorry, don't know how to use latex at the moment, but:

Summation from k=0 to infinity of (k+2)2^(-k)

$\sum (k+2)2^{-k}$

It says to sum the series.

I'm really unsure where to start in this problem. The only similarity I can notice is 1/(1-x) which equals the summation of x^k, but I don't think you can plug in -k's to change summations, can you?

Any ideas?

Edit-
Nvm, put it in LaTex too.
Is...

$\displaystyle \sum_{k=0}^{\infty} k\ x^{k} = x\ \sum_{k=1}^{\infty} k\ x^{k-1} = x\ \frac{d}{dx} \sum_{k=0}^{\infty} x^{k}=$

$\displaystyle = x\ \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^{2}}$ (1)

... and...

$\displaystyle 2\ \sum_{k=0}^{\infty} (\frac{x}{2})^{k} = \frac{4}{2-x}$ (2)

... so that the sum of the series is the sum of (1) for $x=\frac{1}{2}$ and (2) for $x=1$...

$\displaystyle \sum_{k=0}^{\infty} \frac{k+2}{2^{k}} = 2+4=6$ (3)

Merry Christmas from Italy

$\chi$ $\sigma$