1) 1 = 1 2) 2 + 3 + 4 = 1 + 8 3) 5 + 6 + 7 + 8 + 9 = 8 + 27 4) 10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64 = 91 Each left side is an arithmetic series 1. First we make a conjecture about the first term of each arithmetic series. Those first terms on the left go 1,2,5,10. Let's compare that to a list of squares beginning with 0², i.e., 1.2.4.9: n 1,2,3, 4 an 1,2,5,10 (n-1)² 0,1,4, 9 each number is square of 1 more than the square of 1 less than n so the general term is (n-1)²+1, or n²-2n+1+1or n²-2n+2. n²-2n+2 2. Next we make a conjecture about the number of terms in each arithmetic series. The number of terms on the left go 1,3,5,7 which is the odd numbers. We compare them to the even integers, 2n n 1,2,3,4 an 1,3,5,7 2n 2,4,6,8 We see that each is one less than 2n. So the general term for the number of terms is 2n-1 3. Next we make a conjecture about the common difference of the terms of each arithmetic series on the left. That's easy. The difference is alsways 1. The formula for the sum of an arithmetic series is Sn =[2a1 + (n-1)d] except in place of n on the right side, we use the general expression for the number of term. So we replace n by 2n-1. We replace a1 by the general expression for the first term, n²-2n+2, and of course d by 1: Sn = [2(n²-2n+2) + ({2n-1}-1)1], and simplify: Sn = [2n²-4n+4 + (2n-1-1)], Sn = [2n²-4n+4+2n-1-1], Sn = [2n²-2n+2], Sn = [2(n²-n+1)], Sn = (2n-1)(n²-n+1) So (2n-1)(n²-n+1) is the general term for the left side. Now we look at the right sides. All of them have two terms except the first, but it can be considered as also having two terms 0 + 1. So we have 0 + 1, 1 + 8, 8 + 27, 27 + 64. We recognize those as cubes: 0³+1³, 1³+2³, 2³+ 3³, 3³ + 8³ So the general term on the right is (n-1)³+n³ We can factor that as the sum of two cubes: (n-1)³+n³ = [(n-1)+n]{(n-1)²-(n-1)n+n²] = (n-1+n)(n²-2n+1-n²+n+n²) = (2n-1)(n²-n+1) That's the same general term we got for the left side. So we have proved the conjecture, that in each case the left side will equal the right side. Edwin