Proof By Induction Trouble

Q1.

Consider the sum to (n+1). = [Sum to n] + (m^2) where m = n+1

= [ n(n+1)(2n+1)/6 ] + (n+1)^2

Multiply and divide the (n+1)^2 by 6 and then expand out of the brackets. You'll find that you can rebracket the numerator now as (n+1)(n+2)(2n+3)

So by manually adding in the (n+1) term, you've found that it does indeed result in the same formula but with all the n replaced by (n+1).

All you then have to do is show that the sum is correct for n=1 (trivial).

This is proof by induction - you show that the formula holds for n=1 and you also show that if it is true for n then it is true for n+1. Hence it is tre for all integers n >= 1.

You can do the same for the other one.

zaza