<s>[INST] Closed form of the series. 
Write the following series in closed form
  $$\frac{1}{a-1}+\frac{2}{a-2}+\frac{2}{a-3}+\frac{1}{a-4}+\frac{1}{a-5}+\frac{2}{a-6}+\frac{2}{a-7}+\frac{1}{a-8}+\frac{1}{a-9}+\frac{2}{a-10}+\frac{2}{a-11}+\cdots+\frac{1}{a-4p+3}+\frac{2}{a-4p+2}+\frac{2}{a-4p+1}+\frac{1}{a-4p},$$
  where $a$ is any positive number.

I tried to split this series into 4 sub series as follows
$$\frac{1}{a-1}+\frac{1}{a-5}+\frac{1}{a-9}+\cdots+\frac{1}{a-4p+3}+$$ 
$$+\frac{2}{a-2}+\frac{2}{a-6}\frac{2}{a-10}+\cdots+\frac{2}{a-4p+2}+$$
$$+\frac{2}{a-3}+\frac{2}{a-7}+\frac{2}{a-11}+\cdots+\frac{2}{a-4p+1}+$$
$$+\frac{1}{a-4}+\frac{1}{a-8}+\frac{1}{a-12}+\cdots+\frac{1}{a-4p}$$
but still unable to obtain the result.
 [/INST] You have splitted the series in $4$ subseries. We see that for each sub series $(a-j)(a-i)=4,\forall j>i$. So we get the following series for every sub series:
$$(1):\sum_{i=0}^{p-1} \frac{1}{a-(4i+1)}\\(2):\sum_{i=0}^{p-1}\frac{2}{a-(4i+2)}\\(3):\sum_{i=0}^{p-1}\frac{2}{a-(4i+3)}\\(4):\sum_{i=0}^{p}\frac{1}{a-4i}.$$ Together in one expression we get:$$\frac{1}{a-4p}+\sum_{i=0}^{p-1} \left(\frac{1}{a-(4i+1)}+\frac{2}{a-(4i+3)}+\frac{2}{a-(4i+2)}+\frac{1}{a-4i}\right)$$
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