<s>[INST] Sum of the following series  I'm stuck with this series:$$\frac12 +  \frac{1\cdot 4}{2 \cdot 5} +  \frac{1\cdot 4 \cdot 7}{2 \cdot 5 \cdot8} +  \frac{1 \cdot 4 \cdot 7 \cdot 10}{2 \cdot 5 \cdot 8 \cdot 11}+\ldots$$
I cant even find the $n$-th th term here... 
I have to prove that   $$\sum_{k=1}^n A_k = \frac12 \left[ \frac{4\cdot 7 \cdot 10 \cdot \ldots \cdot(3n+1)}{2 \cdot 5 \cdot 8 \cdot\ldots\cdot(3n -1)}\  -1 \right] $$  
Here are the instructions given in the question:   write the $A_n$ term of this series  and express $A_{n+1}$ using $A_n$  
find $C$ and $D$ such that  $f(n) = (Cn + D)A_{n+1}$ and  $f(n) - f(n -1) = A_n$
 and after that i have to come up with that above answer...   P.S :- Im stuck in the step of writing the $A_n$ term. I don't know how to write a $n^{th}$ term for a series like this. I think I could manage if someone could help show me how to write the $n^{th}$ term.
 [/INST] Notice that the $n$-th term of the sum is given by:
$$A_n=\prod_{j=1}^{n}\frac{3j-2}{3j-1}.$$
Let now $B_n$ be:
$$ B_n = \prod_{j=1}^{n}\frac{3j+1}{3j-1}=(3n+1)\prod_{j=1}^{n}\frac{3j-2}{3j-1}=(3n+1)A_n=(3n-1)A_n+2A_n.$$
Since:
$$ (3n-1)A_n = (3n-2)A_{n-1} = B_{n-1}$$
it happens that:
$$ B_n = B_{n-1} + 2 A_n = B_{n-2} + 2 A_{n-1} + 2 A_n = \ldots =B_1 + 2\sum_{k=1}^{n}A_k.$$
Since $B_1 = 2$, the last identity gives:
$$\sum_{k=1}^{n}A_k = \frac{B_n-2}{2}.$$
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