Summation: Prove sum[r=1,n][6r^2+32r] = n(n+1)(2n+17)

7a, which you say you have done, says that this sum, up to n, is equal to n(n+1)(2n+ 17). To do 7b, it is sufficient to observe that when n= 1000, 2n+ 17= 2017.
 
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