Summation series

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

please help with these two questions as they seem impossible

YEEEEEET Oct 7, 2018

#1**+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 !!!.

Guest Oct 7, 2018

#2**+2 **

two formulas are of use

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

.Rom Oct 8, 2018