Question

Use mathematical induction to prove that the statement is true for every positive integer n.
8 + 16 + 24 + . . . + 8n = 4n(n + 1)

Answer 1

We can do this just by proving n and n+1. If n=1, then 8=4n(n+1)=4*2=8. n+1 is a tiny bit trickier, but we should still be able to do it. The equation is \[\sum_{1}^{n}8n=4n^2+4n\]So if we're looking to prove the case of n+1, knowing that when n=1 we are correct, we have \[\sum_{1}^{n+1}8n=(\sum_{1}^{n}8n) +8n +8=4(n+1)^2+4(n+1)=\] (distributing the square and the rest of the parenthesis)\[4n^2+8n+4+4n+4=4n^2+12n+8=4n^2+4n+(8n+8)\] Look at the second value in the second equation. That is 8n+8 more than the first equation was, and we've just shown that the polynomial is also 8n+8 more than the first equation. Q.E.D.

Answer 2

Ah! I see! Thanks so much!