<s>[INST] Use induction to prove following sum identity Using induction, find an explicit formula for the sum $\displaystyle\sum_{i=1}^n\frac{3}{i^2+3i}.$
My attempt: After writing the the first few partial sums, I was able to tell that the denominators of the sum will have the form of $(i+3)!/6$ but I'm not sure how to proceed or find a nice pattern for the numerators: 3,21,146,1022,...
Any help would be appreciated.
Edit: Use Induction, not telescoping. I know how to telescope, what I am looking for is the explicit formula of the sum without using telescoping.
 [/INST] It looks like a telescoping series to me.  $\frac{3}{i(i+3)}=\frac{1}{i}-\frac{1}{i+3}$, so write out some terms and see what cancels.  It looks like you are left with $$1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}$$
To do induction, claim $S_N=1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}$.  Show that it works when $N=1$ (base case). 
Now prove that $$S_{N+1}=\left(\sum_{i=1}^{N}\frac{3}{i^2+3i}\right) + \frac{3}{(N+1)^2+3(N+1)}$$
And this is $$\left(1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}\right)+\frac{3}{(N+1)^2+3(N+1)}$$
using the induction hypothesis.
Show this adds up to $$S_{N+1}= 1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{N+2}- \frac{1}{N+3}- \frac{1}{N+4}$$ which is your formula you are trying to prove.
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