Summation $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$

Question

I came across a question today...

Q. The sum $\dfrac{3}{1^2}+\dfrac{5}{1^2+2^2}+\dfrac{7}{1^2+2^2+3^2}+....$ upto $11$ terms is?

Okay, I think it can be written as

$$\sum_{r=1}^{11}\dfrac{2r+1}{1^2+2^2+...+r^2}$$

I can't see how to simplify it.

Another way I tried is: $$\dfrac{2^2-1^2}{1^2}+\dfrac{3^2-2^2}{1^2+2^2}+\dfrac{4^2-3^2}{1^2+2^2+3^2}+....$$

I simplified it to

$$-1+2^2(\dfrac{1}{1^2}-\dfrac{1}{1^2+2^2})+3^2(\dfrac{1}{1^2+2^2}-\dfrac{1}{1^2+2^2+3^2})+....$$

$$\Rightarrow -1+\dfrac{2^4}{1^2(1^2+2^2)}+\dfrac{3^4}{(1^2+2^2)(1^2+2^2+3^2)}+....$$

I can't see any way after it too... :(

Can you please help?

Answer 1

As $1^2+2^2+\cdots+r^2=\dfrac{r(r+1)(2r+1)}6$

$$\dfrac{2r+1}{1^2+2^2+\cdots+r^2}=6\left(\dfrac{r+1-r}{r(r+1)}\right)=?$$

Can you recognize the Telescoping nature?

Answer 2

After the help provided by @lab bhattacharjee, I would like to complete the answer.

$$6\sum_{r=1}^{11}\left(\dfrac{r+1-r}{r(r+1)}\right)$$

$$=6\sum_{r=1}^{11}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right)$$

$$=6\left[ \left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+....+\left(\dfrac{1}{11}-\dfrac{1}{12}\right) \right]$$

$$=6\left(\dfrac{1}{1}-\dfrac{1}{12}\right)$$

$$=\dfrac{11}{2}$$