Simple induction algebra

**Jones** · October 14th 2009, 12:34 PM

Hi,

I were reviewing some induction and saw this:

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

And the solution looks like this:

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

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

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

**Raoh** · October 14th 2009, 01:12 PM

hi

$\sum_{k=1}^{n+1}\frac{k^{2}}{2^{k}}=6-\frac{n^2+4n+6}{2^{n}}+\frac{(n+1)^2}{2^{n+1}}$ , proceed like this

(from where they got the $r$ ?

)

**Jones** · October 14th 2009, 03:23 PM

Originally Posted by Raoh

hi

$\sum_{k=1}^{n+1}\frac{k^{2}}{2^{k}}=6-\frac{n^2+4n+6}{2^{n}}+\frac{(n+1)^2}{2^{n+1}}$ , proceed like this

(from where they got the $r$ ?

)

Sorry, the 'r's are supposed to be n's

**Raoh** · October 15th 2009, 03:46 AM

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

**Jones** · October 15th 2009, 06:24 AM

Originally Posted by Raoh

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

Hi,

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

**Raoh** · October 15th 2009, 07:05 AM

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

**Jones** · October 15th 2009, 07:08 AM

Originally Posted by Raoh

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

But they have different denominators?

**Raoh** · October 15th 2009, 07:16 AM

yes

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

**Jones** · October 15th 2009, 11:07 AM

Originally Posted by Raoh

yes

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

neat!

Thanks!

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