My Math Forum Saw these at another forum

 Number Theory Number Theory Math Forum

 October 7th, 2018, 02:19 PM #1 Senior Member     Joined: Sep 2015 From: USA Posts: 2,195 Thanks: 1152 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, 06:18 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,950 Thanks: 1842 (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, 10:22 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 270 Thanks: 31 Can anyone post the solution of the first problem? I never saw such a problem Last edited by skipjack; October 9th, 2018 at 01:15 AM.
 October 8th, 2018, 11:33 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,508 Thanks: 2513 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 11:37 AM.
 October 8th, 2018, 08:06 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,508 Thanks: 2513 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 08:22 PM.
 October 9th, 2018, 01:27 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,950 Thanks: 1842 (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 Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Rocket88 New Users 5 May 13th, 2016 12:20 PM spo0ofer Probability and Statistics 1 April 18th, 2016 04:02 PM Tores1 Pre-Calculus 4 November 2nd, 2015 04:42 AM Mark W. New Users 6 September 16th, 2008 09:15 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top