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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 $3\sum \limits_{r=1}^n~(r2)(r3)= (n1)(n2)(n3)+6$ The second one could probably be proven by brute force but I'm sure you number theory folks have shortcuts for these. 
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}
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~(r2)(r3) &= 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^215n + 36n) \\ &= \tfrac12(2n^3  12n^2 + 22n) \\ &= n^3  6n^2 + 11n  6 + 6 \\ &= (n1)(n^2  5n + 6) + 6 \\ &= (n1)(n2)(n3) + 6 \end{align} \begin{align}3\sum \limits_{r=1}^n~(r2)(r3) &= 3\sum_{r=2}^{n3}(r+1)r \\ &= 6 + 0 + 0 + 3\sum_{r=0}^{n3}(r+1)r \\ &= 6 + 3\sum_{r=0}^{n3}(r^2+r) \\ &= 6 + \tfrac36(n3)\big(2(n3)+1\big)\big((n3)+1\big) + \tfrac32(n3)\big((n3)+1\big) \\ &= 6 + \tfrac12(n3)(2n5)(n2) + \tfrac32(n3)(n2) \\ &= 6 + (n3)(n2)\left(\tfrac12(2n5) + \tfrac32\right) \\ &= 6 + (n3)(n2)(n1) \\ \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. 

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