June 4th, 2014, 03:25 AM | #1 |
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8 | sum2س
sum 1/(1*2*3*4) +4/(2*4*5*6)+9/(5*6*7*8)+.................. Last edited by skipjack; June 4th, 2014 at 06:00 AM. |
June 4th, 2014, 04:34 AM | #2 |
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra |
$$\frac{n^2}{n(n+1)(n+2)(n+3)}$$ $\displaystyle =\ \frac{-1}{2(n+1)} + \frac2{n+2} + \frac{-3}{2(n+3)}$ $\displaystyle =\ -\frac12\left(\frac1{n+1}-\frac1{n+2}\right) + \frac32\left(\frac1{n+2}-\frac1{n+3}\right)$ Start telescoping. Last edited by Olinguito; June 4th, 2014 at 04:36 AM. |
June 4th, 2014, 05:19 AM | #3 |
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions |
Olinguito, your sequence does not match that of the OP when n=2 or n=3. However, I think Mared meant to put a 3 instead of a 2 on the denominator of the second term $\displaystyle S_n = \frac{1}{1\times2\times3\times4} + \frac{4}{3\times4\times5\times6} + \frac{9}{5\times6\times7\times8} + ...$ $\displaystyle S_n = \sum_1^{\infty}\frac{n^2}{2n(2n-1)(2n+1)(2n+2)} = \sum_1^{\infty}\frac{n^2}{4n(n+1)(2n-1)(2n+1)}$ which can be solved using the same technique that you suggested. |