Joe Peterson secondary student How do I prove by the principal of mathematical induction? 1.n+2.(n-1)+3.(n-2)+.....+(n-2).3+(n-1).2+n.1=(n(n+1)(n+2))/6 Hi Joe, Step1: Suppose n=1 Left side = 1x1 = 1 right side = (1)(1+1)(1+2)/6 = (1x2x3)/6 = 1 So the equation is true for n=1 Step 2: (Inductive step) Suppose the equation is true for n=k. Then: 1k+2(k-1)+3(k-2)+.....+(k-2)3+(k-1)2+(k)1=(k(k+1)(k+2))/6   We want to prove the equation is true for n=(k+1) When n=k+1: L.S. = 1(k+1) + 2(k) + 3(k-1) + 4(k-2) + ... + (k)2 + (k+1)1 Each term (except the last) can be split into two parts: L.S. = 1k+1 + 2(k-1)+2 + 3(k-2)+3 + 4(k-3)+4 + ... + (k)1+k + (k+1)1 Rearrange: L.S. = 1k+2(k-1)+...+(k-1)2+(k)1 + 1+2+3+...+k + (k+1)1 Use the truth of equation for n=k and the known expression for 1+2+3+...+k to rewrite the left side. Now simplify and verify that this is the right side with n = k + 1. Therfore, by induction, the equation is valid for all positive integral values of n. Cheers, Paul Go to Math Central