My Math Forum series

 Algebra Pre-Algebra and Basic Algebra Math Forum

 September 14th, 2009, 04:52 AM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 series If $S=1(n)+2(n-1)+3(n-2)+...+r(n+1-r)+...+n(1)$ and $T=1(n-1)+2(n-2)+3(n-3)+...+r(n-r)+...+(n-1)(1)$ where n is a positive integer, prove that $S+T=\frac{1}{6}n(n+1)(2n+1)$
 September 14th, 2009, 10:49 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 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)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:
 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)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.
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: 20,469 Thanks: 2038 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:
 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 + 2² + . . . + n²        = 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: 20,469 Thanks: 2038 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.

 Tags series

 Thread Tools Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top