September 14th, 2009, 04:52 AM  #1 
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  series
If and where n is a positive integer, prove that 
September 14th, 2009, 10:49 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
Since S and T are functions of n, I will denote them as S(n) and T(n) respectively. From the given definitions, S(k) = T(k + 1), where k is any positive integer. Let U(k) = S(k) + T(k) and V(k) = k(k + 1)(2k + 1)/6, where k is any positive integer, so that U(1) = 1 + 0 = 1, and V(1) = 1. U(k + 1)  U(k) = S(k+1) + T(k+1)  (S(k) + T(k)) = S(k + 1)  T(k) = 1(k+1) + 2(k) + 3(k1) + ... + r(k+2r) + (r+1)(k+1r) ... + (k1)(3) + k(2) + (k+1)(1)  1(k1)  2(k2)  3(k3)  ...  r(kr)  ...  (k2)(2)  (k1)(1) = (k + 1) + (k + 1) + (k + 1) + ... + (k + 1) + (k + 1) (k + 1 times in total) = (k + 1)². V(k + 1)  V(k) = (k + 1)(k + 2)(2(k + 1) + 1)/6  k(k + 1)(2k + 1)/6 = (k + 1)(2k² + 7k + 6  (2k² + k))/6 = (k + 1)(6k + 6)/6 = (k + 1)². Hence U(k) ? V(k) by induction. 
September 15th, 2009, 03:05 AM  #3  
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: Quote:
S+T=1(n)+3(n1)+5(n2)+7(n3)+...+2r(2n2r+1)+...+(2n1) how can I continue from here? Thanks !!!  
September 15th, 2009, 04:54 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
Instead, why not use the following? U(n) = U(1) + U(2)  U(1) + U(3)  U(2) + . . . + U(n)  U(n  1) = 1 + 2² + . . . + n² = n(n + 1)(2n + 1)/6 
September 16th, 2009, 07:17 AM  #5  
Senior Member Joined: Sep 2008 Posts: 199 Thanks: 0  Re: Quote:
 
September 16th, 2009, 07:47 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
Yes, I defined U as S + T. If you wish, you can also do V(n) = V(1) + V(2)  V(1) + . . . + V(n)  V(n  1) = 1 + 2² + . . . + n². 

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