How to find the sum of the following series from $n=0$ to $n=99$ Find the summation of the series from $n=0$ to $n=99$. The question was given in the format of 
$$(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100).$$
I was able to generalise it but could not solve it. Help!! the general formula is summation (2n+2)!/(2n)!

We have that
$$n(n-1)=\frac{n^3-(n-1)^3}{3}-\frac{1}{3}.$$
Therefore
$$\begin{align}
(1\cdot 2)+(3\cdot 4)+(5\cdot 6)+\dots +(99\cdot 100)&=\sum_{n=1}^{50}(2n-1)(2n)=4\sum_{n=1}^{50}n(n-1)+2\sum_{n=1}^{50}n\\
&=4\sum_{n=1}^{50}\frac{n^3-(n-1)^3}{3}-\frac{4\cdot 50}{3}+50\cdot 51\\
&=\frac{4}{3}\left(50^3-0^3\right)-\frac{200}{3}+50\cdot 51=169150
\end{align}$$
where we noted that the last sum on the right is telescopic.
