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Summation series 

https://snag.gy/3A9Ct4.jpg

please help with these two questions as they seem impossible

 Oct 7, 2018
 #1
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+1

Sorry, I can't see that the sum of Sigma is a multiple of 2017, which is a prime number !! Even though they ask to prove it without summing it up, I still summed it up and got 1,986,985,000 which is no multiple of 2017. In order for the sum to be a multiple of 2017, it has to sum up to 1,986,985,023 !!!.

 Oct 7, 2018
 #3
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Yes, it sums up to what Rom got! Your mistake was summing up 6r^2 - 32r. It is 6r^2 + 32r !!.

Guest Oct 8, 2018
 #2
avatar+4827 
+2

two formulas are of use

 

\(\sum \limits_{r=1}^n~r^2 = \dfrac{1}{6} n (n+1) (2 n+1) \\ \sum \limits_{r=1}^n~r = \dfrac 1 2 n(n+1) \\ \text{applying these we get, after some algebra}\\ \sum \limits_{r=1}^n~6r^2 + 32r = n (n+1) (2 n+17) \\ \text{plugging }n=1000 \text{ into the rightmost factor we see that }\\ 2(1000)+17 = 2017 \\ \text{and thus the overall sum is a multiple of 2017}\)

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 Oct 8, 2018

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