Page 57 - 'Blast_Into_Math

Blast into Math! ets of nummers: mathematical plaagrounds

so we can substitute:

n(n +1)
1+2+ ... + n +(n +1)= +(n +1).
2

We can combine these numbers into a single fraction by putting the multiplicative identity in

disguise:

2
1= ,
2

so we can use this to put n +1 into disguise

2 2(n +1)
n +1 =1 ∗ (n +1)= ∗ (n +1)=
2 2

so

n(n +1) n(n +1)+ 2(n +1)
1+2+ ... + n +(n +1)= +(n +1)= .
2 2

Look carefully. We can re-arrange the numerator

n(n +1)+ 2(n +1) (n +1)(n +2)
= .
2 2

So, we have shown that

(n +1)(n +2)
1+2+ ... + n +(n +1)= .
2

This is the formula for n +1, so we have proven the final step of induction: if the theorem is
true for n , then it is true for n +1.

Induction proofs may seem difficult at first, but please don’t get discouraged. If you carefully and patiently

work through all the induction proofs at the end of this chapter, you’ll have your very own mathemagical
induction escalator which can bring you to infinite heights!

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