1. ## Simple induction algebra

Hi,

I were reviewing some induction and saw this:

prove that for all integers n:
$\displaystyle \sum_{k=1}^n \frac{k^2}{2^k} = 6-\frac{n^2+4n+6}{2^n}$

And the solution looks like this:

$\displaystyle \sum_{k=1}^{n+1} \frac{k^2}{2^k}6-~\frac{n^2+4n+6}{2^n} + \frac{(n+1)^2}{2^{n+1}} \longrightarrow$

$\displaystyle \frac{2n^2+8r+12-r^2-2r-1}{2^{r+1}}=...$

Whats going on here? why did they subtract $\displaystyle r^2+2r+1$
And the reason they multiplied the expression by 2 where to get the same denominator, but the denominator in the new expression is still $\displaystyle 2^{n+1}$ why?

2. hi
$\displaystyle \sum_{k=1}^{n+1}\frac{k^{2}}{2^{k}}=6-\frac{n^2+4n+6}{2^{n}}+\frac{(n+1)^2}{2^{n+1}}$, proceed like this (from where they got the $\displaystyle r$ ? )

3. Originally Posted by Raoh
hi
$\displaystyle \sum_{k=1}^{n+1}\frac{k^{2}}{2^{k}}=6-\frac{n^2+4n+6}{2^{n}}+\frac{(n+1)^2}{2^{n+1}}$, proceed like this (from where they got the $\displaystyle r$ ? )
Sorry, the 'r's are supposed to be n's

4. try to expand what i gave you,you should end up with something like this,
$\displaystyle \frac{6(2^n+1)-(n+1)^2-4(n+1)-6}{2^{n+1} }$

5. Originally Posted by Raoh
try to expand what i gave you,you should end up with something like this,
$\displaystyle \frac{6(2^n+1)-(n+1)^2-4(n+1)-6}{2^{n+1} }$
Hi,

Hm, to get the whole expression over $\displaystyle 2^{n+1}$ you multiplied by 2 ?

6. i just added the term $\displaystyle \left [\frac{(n+1)^2}{2^{n+1}} \right ]$.

7. Originally Posted by Raoh
i just added the term $\displaystyle \left [\frac{(n+1)^2}{2^{n+1}} \right ]$.
But they have different denominators?

8. yes,but you have $\displaystyle 6-\frac{n^2+4n+6}{2^{n}} = \frac{6\times2^n-n^2-4n-6}{2^n}=\frac{2(6\times2^n-n^2-4n-6)}{2^{n+1}}$.

9. Originally Posted by Raoh
yes,but you have $\displaystyle 6-\frac{n^2+4n+6}{2^{n}} = \frac{6\times2^n-n^2-4n-6}{2^n}=\frac{2(6\times2^n-n^2-4n-6)}{2^{n+1}}$.
neat!

Thanks!