How do i factor that induction expression?

[James889]

Hi,

i need to prove that $$\sum^n_{k=1}\frac{k^2}{2^k} = 6-~\frac{n^2+4n+6}{2^n}~~(*)$$

for n=1 the statement holds as $$\sum\frac{1}{2} = 6 -\frac{11}{2}$$

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

In order to get a common denominator multiply (*) by 2

$$\frac{2n^2 + 8n +12 + n^2 +2n +1}{2^n} \longrightarrow \frac{3n^2 + 10n +13}{2^n}$$

But how do i factor that expression ?

 

[HallsofIvy]

Hi,

i need to prove that $$\sum^n_{k=1}\frac{k^2}{2^k} = 6-~\frac{n^2+4n+6}{2^n}~~(*)$$

for n=1 the statement holds as $$\sum\frac{1}{2} = 6 -\frac{11}{2}$$

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

You have a mistake here. Your sum adds numbers of the form $k^2/2^n$ so the term added should be
$$\frac{(n+1)^2}{2^n}$$

In order to get a common denominator multiply (*) by 2

$$\frac{2n^2 + 8n +12 + n^2 +2n +1}{2^n} \longrightarrow \frac{3n^2 + 10n +13}{2^n}$$

But how do i factor that expression ?

 

[James889]

okay,

So now we have $$\frac{2n^2+6n+7}{2^n}$$

But shouldn't the new expression be of the form $$\frac{\text{polynomial}}{2^{n+1}}$$?

 

[lanedance]

i haven't checked your answer, but to get in the form you want, you could you try multiplying by 2/2 then rearranging the polynomial in terms of (n+1)