Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$ $$
\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots
$$
up to $n$ terms. I need help in solving this sum. I tried finding the coefficients of terms after splitting the terms..: it becomes
$$(\frac{1}{1\cdot 6}-\frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}-\frac{1}{6\cdot4}) + (\frac{1}{6\cdot 2} - \frac{1}{3\cdot2} +\frac{1}{4\cdot 2} -\frac{1}{6\cdot5})+\cdots.$$ I tried solving it but am getting nowhere .Someone please help me with this sum.

We can use the following identity: $$\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left(\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right).$$
Thanks to this identity, if we want to compute $$\sum_{k=1}^n\frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3}\sum_{k=1}^n\left(\frac{1}{k(k+1)(k+2)}-\frac{1}{(k+1)(k+2)(k+3)}\right),$$
we only have to subtract $1/(n+1)(n+2)(n+3)$ from $1/(1\cdot2\cdot3)$ and then devide it by 3, because all other terms cancel out. This gives us the result: $$\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}.$$
