<s>[INST] 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.
 [/INST] HINT:
The $r(\ge1)$th term
$$=\dfrac1{r(r+1)(r+2)(r+3)} =\dfrac{r+3-r}{3r(r+1)(r+2)(r+3)}=v_r-v_{r+1}$$
where $v_m=\dfrac1{3m(m+1)(m+2)}$
$$\sum_{r=1}^n u_r=\sum_{r=1}^n(v_r-v_{r+1})=v_1-v_{n+1}$$
Can you take  it from here?
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