Prove by induction that, for all integers n >=1 , ∑(from r=1 to n) r(2r−1)(3r−1)=(n/6)(n+1)(9n^2 -n−2). Assume that 9(k+1)^2 -(k+1)-2=9k^2 +17k+6

First, recall how to construct a proof by simple induction in this manner: (1) Assume statement true for n=k, (2) Prove true for n=k+1, (3) Show true for n=1.(1) => ∑(from r=1 to k) r(2r−1)(3r−1)= (k/6)(k+1)(9k^2 -k−2).(2) To prove true for n=k+1, first acknowledge that ∑(from r=1 to k+1)= ∑(from r=1 to k) + (k+1)th term of the equation in r.=> ∑(from r=1 to k+1) r(2r−1)(3r−1)= (k/6)(k+1)(9k^2 -k−2) {sum from r=1 to k} + (k+1)(2(k+1)-1)(3(k+1)-1) {(k+1)th term}.Bringing out the common term of the two products, =(k+1)/6 [(k)(9k^2 -k-2) + 6(2k+1)(3k+2)] Rearranging, =(k+1)/6 [9k^3 -k^2 +36k^2 -2k+24k+18k+12] =(k+1)/6 [9k^3 +35k^2 +40k+12].Factorising a (k+2) term, =((k+1)/6)(k+2) [9k^2 +17k+6].Using information given in the question, =((k+1)/6)(k+2) [9(k+1)^2 -(k+1)-2] ; which is in exactly the correct form as required given that n=k+1 THEREFORE proven true for n=k+1.(3) When n=1, LHS= 1(2(1)-1)(3(1)-1) = 2; RHS= (1/6)(1+1)(9(1)^2 -1-2) = 2 THEREFORE shown that true for n=1.Hence, we have written a proof by induction showing that ∑(from r=1 to n) r(2r−1)(3r−1)=(n/6)(n+1)(9n^2 -n−2) is true for all integers greater than or equal to 1.