# Finding the solution of a summation

The summation is as follows: \begin{eqnarray*} \sum_{i=1}^n\frac{2+i}{2^i} \end{eqnarray*} For some reason, I can't seem to figure out what I'm supposed to do here. I tried breaking it up like \begin{eqnarray*} \sum_{i=1}^n\frac{2}{2^i}+\sum_{i=1}^n\frac{i}{2^i} \end{eqnarray*} but I still don't know what I can do with the closed form.

I tried finding a pattern as well by plugging and chugging, and I came up with sum from $i = 3$ to $n$ \begin{eqnarray*} \sum_{i=3}^n\frac{i}{2^i} \end{eqnarray*} but that's just the second half of my first equation.

So confused as to how I might go about solving this one. How would you break this apart to make it easier to work with?

HINT:

$$x\frac{d}{dx}\sum_{i=1}^nx^{i}=\sum_{i=1}^nix^{i}$$

Now, sum the geometric series and let $x=1/2$.

hint

use

$$x+x^2+...x^n=x\frac {1-x^n}{1-x}$$

$$x+2x^2+3x^3+...nx^n=$$

$$x\frac {d}{dx}(x+x^2+....x^n)$$

$$=\frac {x}{(1-x)^2}(nx^{n-1}(x-1)+1-x^n).$$

with $x =\frac {1}{2}$.

Breaking it up is a good idea. You have $$\sum_{i=1}^n \frac{2}{2^i} + \sum_{i=1}^n \frac{i}{2^i}$$ With $x = \frac12$, the values of these summations become clearer: $$2 \sum_{i=1}^n x^i + \sum_{i=1}^n i x^i$$ Now we do a little trick: $ix^i$ is $x$ times the derivative of $x^i$. So we get $$2 \sum_{i=1}^n x^i + x \frac{d}{dx} \sum_{i=1}^n x^i$$ Finally, the standard identity for geometric sums gives you $$2 \frac{x - x^{n+1}}{1 - x} + x \frac{d}{dx} \left( \frac{x - x^{n+1}}{1 -x}\right),$$ and you can now take the derivative of that expression, and then plug in $x = \frac12$.

First note that \eqalign{ & \sum\limits_{1\, \le \,k\, \le \,n} {{k \over {x^{\,k} }}} = \sum\limits_{1\, \le \,k\, \le \,n} {k\left( {{1 \over x}} \right)} ^{\,k} = \sum\limits_{1\, \le \,k\, \le \,n} {k\,y^{\,k} } = \cr & = y{d \over {dy}}\sum\limits_{1\, \le \,k\, \le \,n} {y^{\,k} } = y{d \over {dy}}\left( {y\sum\limits_{0\, \le \,k\, \le \,n - 1} {y^{\,k} } } \right) = y{d \over {dy}}\left( {y{{1 - y^{\,n} } \over {1 - y}}} \right) = \cr & = y\left( {{{\left( {1 - y} \right)\left( {1 - \left( {n + 1} \right)y^{\,n} } \right) + \left( {y - y^{\,n + 1} } \right)} \over {\left( {1 - y} \right)^2 }}} \right) = \cr & = y\left( {{{1 - \left( {n + 1} \right)y^{\,n} + n\,y^{\,n + 1} } \over {\left( {1 - y} \right)^2 }}} \right) = \cr & = {1 \over x}\left( {{{1 - \left( {n + 1} \right)\left( {1/x} \right)^{\,n} + n\,\left( {1/x} \right)^{\,n + 1} } \over {\left( {1 - 1/x} \right)^2 }}} \right) = \cr & = {{x^{\,n + 1} - \left( {n + 1} \right)x + n\,} \over {x^{\,n} \left( {x - 1} \right)^2 }} \cr}

Then it follows easily that \eqalign{ & \sum\limits_{1\, \le \,k\, \le \,n} {{{2 + k} \over {2^{\,k} }}} = \sum\limits_{1\, \le \,k\, \le \,n} {{2 \over {2^{\,k} }}} + \sum\limits_{1\, \le \,k\, \le \,n} {{k \over {2^{\,k} }}} = \cr & = \sum\limits_{1\, \le \,k\, \le \,n} {\left( {{1 \over 2}} \right)} ^{\,k - 1} + \sum\limits_{1\, \le \,k\, \le \,n} {{k \over {2^{\,k} }}} = \cr & = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( {{1 \over 2}} \right)} ^{\,k} + \sum\limits_{1\, \le \,k\, \le \,n} {{k \over {2^{\,k} }}} = \cr & = {{1 - \left( {{1 \over 2}} \right)^{\,n} } \over {1 - \left( {{1 \over 2}} \right)}} + {{2^{\,n + 1} - \left( {n + 1} \right)2 + n\,} \over {2^{\,n} }} = \cr & = 2 - {1 \over {2^{\,n - 1} }} + {{2^{\,n + 1} - 2 - n\,} \over {2^{\,n} }} = {{2^{\,n + 2} - 4 - n\,} \over {2^{\,n} }} \cr}

• Where did the $2$ factor go from the $\frac{2}{2^k}$ go? – Podo Apr 28 '17 at 22:53
• @JeffreyDilley: $2/2^k = 1/2^{k-1}$ then in the following step it becomes $1/2^k$ by shifting the summation range (to have it start from $0$, which gives the standard formula for geometric series). – G Cab Apr 29 '17 at 17:56