How to find the sum of the following series from $n=0$ to $n=99$

Question

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)!

Answer 1

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.

Answer 2

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, $$\sum_{n=1}^{50}2n(2n-1) =4\sum_{n=1}^{50}n^2-2\sum_{n=1}^{50}n =\frac23(50\times 51\times 101)-50\times51=169150. $$

Answer 3

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 $$ 2\sum_{i=2}^{100}\left[\binom{i+1}{3}-\binom{i}{3}\right] =2\binom{101}{3}=\frac{2(101)(100)(99)}{6}=333300. $$

Answer 4

$$ \begin{align} \sum_{k=1}^n(2k-1)2k &=\sum_{k=1}^n\left[8\binom{k}{2}+2\binom{k}{1}\right]\\ &=8\binom{n+1}{3}+2\binom{n+1}{2}\\ &=\frac43(n+1)n(n-1)+(n+1)n\\[3pt] &=\frac{n(n+1)(4n-1)}3 \end{align} $$ Plug in $n=50$ to get $$ \sum_{k=1}^{50}(2k-1)2k=169150 $$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}