Show that:
$$\sum_{r=1}^{n} \frac{5r+4}{r(r+1)(r+2)}=\frac{7n^2+11n}{2(n+1)(n+2)}$$
I'm very lost on this question. I initially tried to identify a certain progression that this function adhered to, but the numerator is linear while the denominator is cubic.
For each $r\in\mathbb{N}$, $$ \frac{5r+4}{r(r+1)(r+2)}=\frac{5r+2}{r(r+1)}-\frac{5(r+1)+2}{(r+1)(r+2)}, $$ and therefore \begin{align} \sum_{r=1}^n\frac{5r+4}{r(r+1)(r+2)}&=\frac{5+4}{1\times2\times3}-\frac{5(n+1)+2}{(n+1)(n+2)}\\&=\frac{7n^2+11n}{2(n+1)(n+2)}.\end{align}
We can write $$\frac{5r+4}{r(r+1)(r+2)} = \frac{2}{r} + \frac{1}{r+1} - \frac{3}{r+2}$$ for all $r \in \mathbb{N}$. Considering triplets of consecutive terms in the sum and noticing vanishing expressions, we obtain $$\sum_{r=1}^{n} \frac{5r+4}{r(r+1)(r+2)} = \frac{2}{1} + \frac{1}{2} + \frac{2}{2} - \frac{3}{n+1} + \frac{1}{n+1} - \frac{3}{n+2}$$ which yields the desired result.
This can be easily proven using Induction.
Testing for $n=1$, it is true. Let it be true for $n=k$. Then, $$\sum_{r=1}^k \frac{5r+4}{r(r+1)(r+2)} = \frac{7k^2+11k}{2(k+1)(k+2)}$$ Let us try to prove it by induction. For $n=k+1$ $$\sum_{r=1}^{k+1} \frac{5r+4}{r(r+1)(r+2)} = \frac{7k^2+11k}{2(k+1)(k+2)}+\frac{5(k+1)+4}{(k+1)(k+2)(k+3)}=\frac{(7k^2+11k)(k+3)+10k+18}{2(k+1)(k+2)(k+3)}$$ On simplifying it further, we get $$\frac{7k^2+25k+18}{2(k+2)(k+3)} = \frac{7(k+1)^2+11(k+1)}{2(k+2)(k+3)}$$ which is the required form. Hence, it is proved by induction.
