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
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