Joe Peterson
How do I prove by the principal of mathematical induction? Step1: Left side = 1x1 = 1 right side = (1)(1+1)(1+2)/6 = (1x2x3)/6 = 1 So the equation is true for n=1 1k+2(k1)+3(k2)+.....+(k2)3+(k1)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: Each term (except the last) can be split into two parts: L.S. = 1k+1 + 2(k1)+2 + 3(k2)+3 + 4(k3)+4 + ... + (k)1+k + (k+1)1 Rearrange: L.S. = 1k+2(k1)+...+(k1)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. Paul
