If $$S=\sum_{k=1}^{n}\frac{(k^2)(4^k)}{(k+1)(k+2)}$$,then $S$ equals
$$S=\sum_{k=1}^{n}\frac{(k^2-1+1)(4^k)}{(k+1)(k+2)}$$ $$S=\sum_{k=1}^{n} \frac{(k -1)(4^k)}{(k+2)} +\sum_{k=1}^{n}\frac{(4^k)}{(k+1)(k+2)} $$ $$S=\sum_{k=1}^{n} 4^k\left(3-\frac{1}{k+2}\right) +\sum_{k=1}^{n}\frac{(2^{2k+1}-2^{2k})}{(k+1)(k+2)}$$ I tried to covert it into telescoping series but wasn't able to convert it .
Note that $$ S_n=\sum_{k=1}^{n}\frac{k^2 4^{k}}{(k+1)(k+2)}=\sum_{k=1}^{n}\left(1-\frac{3k+2}{(k+1)(k+2)}\right)4^{k}=\sum_{k=1}^{n}\left(1-\frac{4}{k+2}+\frac{1}{k+1}\right)4^{k}\\=\sum_{k=1}^{n}4^k+\sum_{k=1}^{n}\left(\frac{1}{k+1}-\frac{4}{k+2}\right)4^k=\sum_{k=1}^{n}4^k+\sum_{k=1}^{n}\left(\frac{4^k}{k+1}-\frac{4^{k+1}}{k+2}\right); $$ the first term is simple to evaluate, and the second sum is telescoping.