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)!
 A: 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.
A: What you have there is not $\sum_{n=0}^{99}n(n+1)$ but rather
$\sum_{n=1}^{50}2n(2n-1)$. Using standard formulae,
$$\sum_{n=1}^{50}2n(2n-1)
=4\sum_{n=1}^{50}n^2-2\sum_{n=1}^{50}n
=\frac23(50\times 51\times 101)-50\times51=169150.
$$
A: $$
\begin{align}
\sum_{k=1}^n(2k-1)2k
&=\sum_{k=1}^n\left[8\binom{k}{2}+2\binom{k}{1}\right]\\
&=8\binom{n+1}{3}+2\binom{n+1}{2}\\
&=\frac43(n+1)n(n-1)+(n+1)n\\[3pt]
&=\frac{n(n+1)(4n-1)}3
\end{align}
$$
Plug in $n=50$ to get
$$
\sum_{k=1}^{50}(2k-1)2k=169150
$$
A: Observe that your sum may be written as 
$$
\sum_{n=0}^{99}n(n+1)=2\sum_{n=0}^{99}\binom{n+1}{2}=2\sum_{i=2}^{100}\binom{i}{2}.
$$
Moreover, using Pascal's identity and telescoping, we obtain
$$
2\sum_{i=2}^{100}\left[\binom{i+1}{3}-\binom{i}{3}\right]
=2\binom{101}{3}=\frac{2(101)(100)(99)}{6}=333300.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 1}^{50}\pars{2k - 1}2k & =
\left.\partiald[2]{}{x}\sum_{k = 1}^{50}x^{2k}\right\vert_{\ x\ =\ 1} =
\partiald[2]{}{x}\bracks{x^{2}\,{x^{100} - 1 \over x^{2} - 1}}_{\ x\ =\ 1} =
\bbx{169150}
\end{align}
