User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 October 7th, 2018, 01:19 PM #1 Senior Member   Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 Saw these at another forum i) Without calculating $S$ show $S = \sum \limits_{r=1}^{1000}~(6r^2 + 32r) \text{ is a multiple of 2017}$ ii) Prove that $3\sum \limits_{r=1}^n~(r-2)(r-3)= (n-1)(n-2)(n-3)+6$ The second one could probably be proven by brute force but I'm sure you number theory folks have shortcuts for these. Thanks from Joppy October 7th, 2018, 05:18 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,929 Thanks: 2205 (i) Obtain a formula for the first n terms of the sum, then show that one factor of it evaluates to 2017 for n = 1000. (ii) Use mathematical induction. October 8th, 2018, 09:22 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 600 Thanks: 87 Can anyone post the solution of the first problem? I never saw such a problem Last edited by skipjack; October 9th, 2018 at 12:15 AM. October 8th, 2018, 10:33 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra \begin{align}\sum_{r=1}^{n}r^2 &= \frac16 n(2n+1)(n+1) \\ \sum_{r=1}^{n}r &= \frac12 n(n+1) \\[12pt] \sum_{r=1}^{n}(6r^2 + 32r) &= n(2n+1)(n+1) + 16n(n+1) \\ &= (2n+17)n(n+1) \\ \sum_{r=1}^{1000}(6r^2 + 32r) &= 2017 \times 1000 \times 1001\end{align} Thanks from greg1313 Last edited by v8archie; October 8th, 2018 at 10:37 AM. October 8th, 2018, 07:06 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra \begin{align}3\sum \limits_{r=1}^n~(r-2)(r-3) &= 3\sum_{r=1}^{n}(r^2 - 5r +6) \\ &= \tfrac12 n(2n+1)(n+1) - \tfrac{15}2 n(n+1) + 18n \\ &= \tfrac12(2n^3 + 3n^2 + n) - \frac{15}2(n^2+n) + \frac12 (36n) \\ &= \tfrac12(2n^3+3n^2+n - 15n^2-15n + 36n) \\ &= \tfrac12(2n^3 - 12n^2 + 22n) \\ &= n^3 - 6n^2 + 11n - 6 + 6 \\ &= (n-1)(n^2 - 5n + 6) + 6 \\ &= (n-1)(n-2)(n-3) + 6 \end{align} \begin{align}3\sum \limits_{r=1}^n~(r-2)(r-3) &= 3\sum_{r=-2}^{n-3}(r+1)r \\ &= 6 + 0 + 0 + 3\sum_{r=0}^{n-3}(r+1)r \\ &= 6 + 3\sum_{r=0}^{n-3}(r^2+r) \\ &= 6 + \tfrac36(n-3)\big(2(n-3)+1\big)\big((n-3)+1\big) + \tfrac32(n-3)\big((n-3)+1\big) \\ &= 6 + \tfrac12(n-3)(2n-5)(n-2) + \tfrac32(n-3)(n-2) \\ &= 6 + (n-3)(n-2)\left(\tfrac12(2n-5) + \tfrac32\right) \\ &= 6 + (n-3)(n-2)(n-1) \\ \end{align} Last edited by v8archie; October 8th, 2018 at 07:22 PM. October 9th, 2018, 12:27 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,929 Thanks: 2205 (ii) Let $f(n)$ denote the given function of n, and note that $f(1) = 6 = 3(1 - 2)(1 - 3)$, as required. \displaystyle \begin{align*}f(n) - f(n - 1) &= (n - 1)(n - 2)(n - 3) + 6 - ((n - 2)(n - 3)(n - 4) + 6) \\ &= (n - 2)(n - 3)(n - 1 - (n - 4)) \\ &= 3(n - 2)(n - 3) \end{align*} The above suffices to prove the given equation by mathematical induction. Thanks from v8archie Tags forum 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 Rocket88 New Users 5 May 13th, 2016 11:20 AM spo0ofer Probability and Statistics 1 April 18th, 2016 03:02 PM Tores1 Pre-Calculus 4 November 2nd, 2015 03:42 AM Mark W. New Users 6 September 16th, 2008 08:15 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      