# Maths – Induction (part 2)

September 10, 2009 Leave a comment

## A better Example

Let’s take a look at another example, this time lets use:

_{n}

**∑ **i^{2} = (n(n+1)(2n+1)) / 6

^{i=0}

** **

Where we have:

_{n}

L(n) = **∑ **i^{2} and R(n) = (n(n+1)(2n+1)) / 6

^{i=0}

For now I will skip the base case. However, do note that when n=0 for the base case, L(0)=R(0), so that the base case is true. As for the step cases, we have:

**∑**i^{2}=**∑**(n+1)^{2}

Once this is done, we then need to move everything next to the ‘∑’ onto the right hand side. So we get this:

- before we move we have:

R(n) = (n(n+1)(2n+1)) / 6 - Once the ‘i’ has be moved over, then R(n) becomes R(n+1), and we have:

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

^{ }

Once we have this, it is just simple arithmetic:

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

(2) = [(n(n+1)(2n+1)) / 6] + [6(n+1)^{2} / 6] *make both have same denominator.*

(3) = (n(n+1)(2n+1) + 6(n+1)^{2}) / 6 *bring both together*

(4) = ((n+1)(n+2)(2n+3)) / 6

(5) = R(n+1)

As from before, step (4) requires you to expand the brackets from step (3), the fractionise the whole lot to obtain**(n+1)(n+2)(2n+3)**.