Stuck in a finite summation in a physics problem.r³/(r-1)² Actually it was not a hard problem as seen from physics point of view, just some application of Coulomb's law (multiple charge system) but it finally ended up as the summation of:- 
$$
\sum_{r=2}^n \frac{(r+1)^3}{r^2}
$$
I tried to open things up but it finally took me to summation of $\frac{1}{r^2}$ and $\frac{1}{r}$, and I know they can be solved but it's too much for my level I guess (as I am a JEE aspirant). Is there any other way to solve the summation?
 A: We have:
$$\frac{(r+1)^3}{r^2}=\frac{r^3+3r^2+3r+1}{r^2}=r+3+\frac{3}{r}+\frac{1}{r^2}$$
Thus:
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\sum_{r=2}^n \bigg( r+3+\frac{3}{r} \frac{1}{r^2} \bigg)$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\sum_{r=2}^n r +\sum_{r=2}^n 3 +3 \sum_{r=2}^n \frac{1}{r} + \sum_{r=2}^n \frac{1}{r^2}$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\bigg( \frac{n(n+1)}{2}-1 \bigg) +3(n-1) +3 \sum_{r=2}^n \frac{1}{r} + \sum_{r=2}^n \frac{1}{r^2}$$
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2}=\frac{n^2+7n-8}{2}+(3H_n-3)+(H_n^{(2)}-1)=\frac{n^2+7n-16}{2}+3H_n+H_n^{(2)}$$
Now, it is not possible to get the exact values for $H_n$ and $H_n^{(2)}$, but for large $n$, you can approximate:
$$\lim_{n \to \infty} H_n - \ln{n} = \gamma = 0.577 \ldots \implies H_n \approx \ln{n}+ \gamma$$
$$\lim_{n \to \infty} H_n - \frac{\pi^2}{6}=0 \implies H_n^{(2)} \approx \frac{\pi^2}{6}$$
Hence, we have:
$$\sum_{r=2}^n \frac{(r+1)^3}{r^2} \approx \frac{n^2+7n-16}{2}+3 \ln{n} + 3 \gamma + \frac{\pi^2}{6}$$
As $n$ gets really large, the difference between the LHS and RHS tends to $0$. This shows that we have a really good approximation.
