Finding the $nth$ term of a sequence Okay so I'm asking this quesion knowing a thing or two about sequences and general terms
What is the sum of the series :
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$

My Try: I tried calculating the general term $T_{n}$ for the sequence but I'm not able to understand how to include every single term of the sequence in a single general term. Writing the series again below the original series with the first term one ahead of the first term of the original series and subtraction didn't help too.How can I get this done? I get a feeling it could be a telescopic sum but unless I have a general term, I can't be sure of it. 

Please suggest me any way of doing this. Thanks for giving this your time.
 A: HINT:
As the number of multiplicand in the numerator is greater than that of denominator and the common differences of numerator & denominator are both $2$
if $$S=1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$
$$\dfrac{-1\cdot S}{2\cdot4}=\dfrac{-1\cdot1}{2\cdot4}+\frac{-1\cdot1\cdot3}{2\cdot4\cdot6}+\frac{-1\cdot1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$$
Now utilize Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
A: $$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$
Numerator is the product of odd positive integers which can be written as
$1 \cdot 3\cdot 5=\dfrac{1 \cdot 2\cdot 3\cdot 4 \cdot 5 }{2  \cdot 4}=\dfrac{5! }{2^2\left(  1\cdot 2\right)}=\dfrac{(2\cdot 2 +1)!}{2^2\cdot 2!}$
$n-$th numerator will be $\dfrac{(2n+1)!}{2^n \,n!}$
denominator is $6\cdot 8$ and can be written as $(2\cdot 3)(2\cdot 4)=2^2\cdot (3\cdot 4)=\dfrac{2^2\cdot 4!}{2}$
next denominator will be $6\cdot 8 \cdot 10=(2\cdot 3)(2\cdot 4)(2\cdot 5)=2^3(3\cdot 4\cdot 5)=\dfrac{2^3\cdot 5!}{2}$
$n-$th denominator will be $2^{n-1} (n+2)!$
Then the $n-$th term for $n\ge 2$ will be
$a_n=\dfrac{\dfrac{(2n+1)!}{2^n \,n!}}{2^{n-1} (n+2)!}=\dfrac{(2n+1)!}{2^n \,n!\,2^{n-1} (n+2)!}=\dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}$
And the series
$$1+\sum _{n=1}^{\infty } \dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}=4$$
