<s>[INST] Summation $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ 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?
 [/INST] After the help provided by @lab bhattacharjee, I would like to complete the answer.
$$6\sum_{r=1}^{11}\left(\dfrac{r+1-r}{r(r+1)}\right)$$
$$=6\sum_{r=1}^{11}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right)$$
$$=6\left[ \left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+....+\left(\dfrac{1}{11}-\dfrac{1}{12}\right) \right]$$
$$=6\left(\dfrac{1}{1}-\dfrac{1}{12}\right)$$
$$=\dfrac{11}{2}$$
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