arrange 1 to 1000 in table w/ 20 rows, 50 cols, so that....

[xwrathbringerx]

a) Is it possible to arrange the integers from 1 to 1000 in a table with 20 rows and 50 columns so that if for each of the rows all the numbers in it are added together, then twenty consecutive integers are obtained?

b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)

 

[soroban]

Re: Random Problems

Hello, xwrathbringerx!

The second one is a "telescopic series".

\(\displaystyle \text{b) Simplify: }\:S \;=\;\frac{1}{2\!\cdot\!5} + \frac{1}{5\!\cdot\!8} + \frac{1}{8\!\cdot\!11} + \cdots + \frac{1}{2000\!\cdot\!2003}\)

\(\displaystyle \text{Note that: }\:\frac{1}{n(n+3)} \;=\;\frac{1}{3}\left(\frac{1}{n} - \frac{1}{n+3}\right)\)


\(\displaystyle \text{Then: }\;S \;=\;\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right) + \frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right) +\left(\frac{1}{8}-\frac{1}{11}\right) + \cdots + \frac{1}{3}\left(\frac{1}{2000} - \frac{1}{2003}\right)\)

. . . . .\(\displaystyle S \;=\;\frac{1}{3}\underbrace{\left(\frac{1}{2} - \frac{1}{5} + \frac{1}{5} -\frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \cdots + \frac{1}{2000} - \frac{1}{2003} \right)}_{\text{Everything cancels except the first and last terms}}\)

. . . . .\(\displaystyle S \;=\;\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2003}\right) \;=\;\frac{1}{3}\left(\frac{2001}{4006}\right) \;=\;\boxed{\frac{667}{4006}}\)