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 October 7th, 2018, 02:19 PM #1 Senior Member     Joined: Sep 2015 From: USA Posts: 2,324 Thanks: 1233 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: 20,310 Thanks: 1978 (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: iPhone Posts: 392 Thanks: 62 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: 7,600 Thanks: 2588 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: 7,600 Thanks: 2588 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: 20,310 Thanks: 1978 (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

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