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 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(k-1) + ... + r(k+2-r) + (r+1)(k+1-r) ... + (k-1)(3) + k(2) + (k+1)(1) � � � � � � � � � � � � � � � � � �- 1(k-1) - 2(k-2) - 3(k-3) - � ... � - r(k-r) - � ... � - (k-2)(2) - (k-1)(1) � � � � � � � � � � � = (k + 1) + (k + 1) + (k + 1) + ... + (k + 1) + (k + 1) � (k + 1 times in total) � � � � � � � � � � � = (k + 1)�. V(k + 1) - V(k) = (k + 1)(k + 2)(2(k + 1) + 1)/6 - k(k + 1)(2k + 1)/6 � � � � � � � � � � � = (k + 1)(2k� + 7k + 6 - (2k� + k))/6 � � � � � � � � � � � = (k + 1)(6k + 6)/6 � � � � � � � � � � � = (k + 1)�. Hence U(k) ? V(k) by induction. September 15th, 2009, 03:05 AM   #3
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 Originally Posted by skipjack 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(k-1) + ... + r(k+2-r) + (r+1)(k+1-r) ... + (k-1)(3) + k(2) + (k+1)(1) � � � � � � � � � � � � � � � � � �- 1(k-1) - 2(k-2) - 3(k-3) - � ... � - r(k-r) - � ... � - (k-2)(2) - (k-1)(1) � � � � � � � � � � � = (k + 1) + (k + 1) + (k + 1) + ... + (k + 1) + (k + 1) � (k + 1 times in total) � � � � � � � � � � � = (k + 1)�. V(k + 1) - V(k) = (k + 1)(k + 2)(2(k + 1) + 1)/6 - k(k + 1)(2k + 1)/6 � � � � � � � � � � � = (k + 1)(2k� + 7k + 6 - (2k� + k))/6 � � � � � � � � � � � = (k + 1)(6k + 6)/6 � � � � � � � � � � � = (k + 1)�. Hence U(k) ? V(k) by induction.
Thanks a lot for the effort, Skipjack, but I haven't learnt mathematical induction, so is it possible to solve this using knowledge on sequences and series? However, this is my attempt ...

S+T=1(n)+3(n-1)+5(n-2)+7(n-3)+...+2r(2n-2r+1)+...+(2n-1)

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 + 2� + . . . + n� � � � �= n(n + 1)(2n + 1)/6 September 16th, 2009, 07:17 AM   #5
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 Originally Posted by skipjack Instead, why not use the following? U(n) = U(1) + U(2) - U(1) + U(3) - U(2) + . . . + U(n) - U(n - 1) � � � �= 1 + 2� + . . . + n� � � � �= n(n + 1)(2n + 1)/6
How did you get this? S+T ?? 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 + 2� + . . . + n�. Tags series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM bilano99 Algebra 4 October 28th, 2011 07:09 AM The Chaz Real Analysis 11 February 7th, 2011 04:52 AM maliniarz Probability and Statistics 1 December 8th, 2010 06:14 PM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

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