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October 7th, 2018, 01:19 PM   #1
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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.
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October 7th, 2018, 05:18 PM   #2
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(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.
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October 8th, 2018, 09:22 AM   #3
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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.
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October 8th, 2018, 10:33 AM   #4
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\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}
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Last edited by v8archie; October 8th, 2018 at 10:37 AM.
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October 8th, 2018, 07:06 PM   #5
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\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.
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October 9th, 2018, 12:27 AM   #6
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(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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