Algebra Pre-Algebra and Basic Algebra Math Forum

 June 4th, 2014, 04:25 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 07:00 AM. June 4th, 2014, 05:34 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 05:36 AM. June 4th, 2014, 06:19 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 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. Thanks from Olinguito Tags sum2س Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

 Contact - Home - Forums - Cryptocurrency Forum - Top      