Series exercise based on raising the exponent to the power of 2!

[wolly]

How does $$\left[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n}{n})^{2}\right]$$ = $$(n-1)a^{2}+2*a\left(\frac{1}{n}+\frac{2}{n}+...+\frac{n-1}{n}\right)+\frac{1^{2}}{n^{2}}+\frac{2^2}{n^{2}}+...+\frac{(n-1)^{2}}{n^{2}}$$?

I tried something but I don't know if this is correct and it's not the same with the upper image!


$$\frac{(an+1)^{2}}{n^{2}}+\frac{(an+2)^{2}}{n^{2}}+\frac{(an+3)^{2}}{n^{2}}+...+\frac{(an+n-1)^{2}}{n^{2}}+\frac{(an+n)^{2}}{n^{2}}$$ = $$\frac{1}{n^{2}}\left((a^{2}n^{2}+2an+1)+(a^{2}n^{2}+4an+4)+(a^{2}n^{2}+6an+9)+...+(a^{2}n^{2}+2an(n-1)+(n-1)^{2}+(a^{2}n^{2}+2an*n+n^{2})\right)$$
$$\frac{(an+1)^{2}}{n^{2}}+\frac{(an+2)^{2}}{n^{2}}+\frac{(an+3)^{2}}{n^{2}}+...+\frac{(an+n-1)^{2}}{n^{2}}+\frac{(an+n)^{2}}{n^{2}}$$

$$\frac{1}{n^{2}}\left((a^{2}n^{2}+2an+1)+(a^{2}n^{2}+4an+4)+(a^{2}n^{2}+6an+9)+...+((a^{2}n^{2}+2an(n-1)+(n-1)^{2})+(a^{2}n^{2}+2an*n+n^{2})\right)$$
Is it correct what I wrote??

 

[blamocur]

How does $$\left[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n}{n})^{2}\right]$$ = $$(n-1)a^{2}+2*a\left(\frac{1}{n}+\frac{2}{n}+...+\frac{n-1}{n}\right)+\frac{1^{2}}{n^{2}}+\frac{2^2}{n^{2}}+...+\frac{(n-1)^{2}}{n^{2}}$$?

I tried something but I don't know if this is correct and it's not the same with the upper image!


$$\frac{(an+1)^{2}}{n^{2}}+\frac{(an+2)^{2}}{n^{2}}+\frac{(an+3)^{2}}{n^{2}}+...+\frac{(an+n-1)^{2}}{n^{2}}+\frac{(an+n)^{2}}{n^{2}}$$ = $$\frac{1}{n^{2}}\left((a^{2}n^{2}+2an+1)+(a^{2}n^{2}+4an+4)+(a^{2}n^{2}+6an+9)+...+(a^{2}n^{2}+2an(n-1)+(n-1)^{2}+(a^{2}n^{2}+2an*n+n^{2})\right)$$
$$\frac{(an+1)^{2}}{n^{2}}+\frac{(an+2)^{2}}{n^{2}}+\frac{(an+3)^{2}}{n^{2}}+...+\frac{(an+n-1)^{2}}{n^{2}}+\frac{(an+n)^{2}}{n^{2}}$$

$$\frac{1}{n^{2}}\left((a^{2}n^{2}+2an+1)+(a^{2}n^{2}+4an+4)+(a^{2}n^{2}+6an+9)+...+((a^{2}n^{2}+2an(n-1)+(n-1)^{2})+(a^{2}n^{2}+2an*n+n^{2})\right)$$
Is it correct what I wrote??

It looks like the right hand side in the first equality is missing a couple of items (typeset in boldface below):
$$(n-1)a^{2}+2*a\left(\frac{1}{n}+\frac{2}{n}+...+\frac{n-1}{n} +\mathbf{\frac{n}{n}} \right) +\frac{1^{2}}{n^{2}}+\frac{2^2}{n^{2}}+...+\frac{(n-1)^{2}}{n^{2}} \mathbf{+\frac{n^2}{n^2}}$$

 

[blamocur]

It looks like the right hand side in the first equality is missing a couple of items (typeset in boldface below):
$$(n-1)a^{2}+2*a\left(\frac{1}{n}+\frac{2}{n}+...+\frac{n-1}{n} +\mathbf{\frac{n}{n}} \right) +\frac{1^{2}}{n^{2}}+\frac{2^2}{n^{2}}+...+\frac{(n-1)^{2}}{n^{2}} \mathbf{+\frac{n^2}{n^2}}$$?

And I missed another one: shouldn't it be $na^2$ instead of $(n-1)a^2$ in the same expression?

 

[blamocur]

Is it correct what I wrote??

It looks correct to me, but you want to group together multiples of $n^2$, multiples of $n$ and "free" elements, i.e., not containing $n$'s.

 

[wolly]

@blamocur And the hint or solution?

 

[wolly]

$$\frac{1}{n^{2}}*\left(n^{2}*a^{2}+n^{2}*a^{2}+....+n^{2}a^{2}\right)$$= $$a^{2}+a^{2}+...+a^{2}$$ and from here?????

 

[blamocur]

and from here?????

$=na^2$, not $(n-1)a^2$ !

 

[blamocur]

@blamocur And the hint or solution?

We don't do solutions here. As for a hint: what have you done with the first hint in my post #4 ?

 

[wolly]

Ok,I found the answer I was looking!