Solving Series Summation Problem: Start & How-To

  • Thread starter Thread starter tykescar
  • Start date Start date
  • Tags Tags
    Series Summation
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
tykescar
Messages
6
Reaction score
0
It isn't homework, it's in a textbook and I'm having trouble with it.

When r=1, summing to n the series of r^3 = (n^2)/4 (n+1)^2

Show that when r = (n+1), summing to 2n = (n^2)/4 (3n+1)(5n+3)

What order do you start the summation, and how do I begin?
 
Physics news on Phys.org
[tex]\sum_{r=1}^{n}r^3=\frac{n^2}{4}\right\left(n+1)\right^2[/tex]

show that

[tex]\sum_{r=n+1}^{2n}r^3=\frac{n^2}{4}\right\left(3n+1)\right\left(5n+3)\right[/tex]

I don't understand how to start
 
Last edited: