# Induction again..... -FP1 OCR

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**Step 1: Show that the solution is true for n=1**

Sum of r^2: (1)^2 = 1

Using sigma notation: 1/6n(n+1)(2n+1)=1/6(1)(2)(3)=1

Therefore we know that the formula works for n=1.

**Step 2: Assume it works for n=k where k is an integer**

If it works for n=k, then sum to k = 1/6k(k+1)(2k+1)

**Step 3: Test for n=k+1 where k is an integer**

Sum to k+1 = sum to k + (k+1)^2 = 1/6k(k+1)(2k+1) + (k+1)^2 = 1/6[k(2k^2+3k+1) + 6k^2+12k+6]

= 1/6[2k^3+3k^2+k+6k^2+12k+6] = 1/6(2k^3+9k^2+13k+6) = 1/6(k+1)(k+2)(2k+3)

Sum to k+1 should equal 1/6(k+1)(k+1+1)[2(k+1)+1] if formula is correct. Because we have managed to show that this is true, we can conclude that the formula works for all integers greater than or equal to 1.

*I know this is confusing, or at least I found it confusing when I did it last year. You really need someone to talk you through out rather than a forum. Look for a video on youtube or see your teacher is my advice :)*Here's a video that might help http://youtu.be/Y1GSE8EZle4

Modified once, last modified by DaveMorris1 on Tue 27th May, 2014 @ 16:40