Question on induction math

0 + 1 + 2 + ... + n + (n+1)
= (0 + 1 + 2 + ... + n) + (n+1)
= ½ × n × (n + 1) + 1 × (n+1) by induction hypothesis
= (½n + 1) × (n + 1)
= (½)(n+1)(n+2)

line 3 puzzles me. It's been a while since I dealt with sums. Does line 3 follow from line 2? By which rule?

Remember (0 + 1 + 2 + ... + n) = ½(n(n+1))
(see page 40 in the textbook)

And if you want to know why that is so:
for 1 ... n you make the following sums:

1 + n = n+1
2 + n-1 = n+1
3 + n-2 = n+1
...
n-2 + 3 = n+1
n-1 + 2 = n+1
n + 1 = n+1

So you have n sums giving you a total of:

n(n+1)

But as you can see, we used (1+n) and all the others twice, so you have to divide by 2:

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

Thanks malberts. I realised that this is a gauss summation.

some summation identities

Gauss summation

i.e. to add:
from 1 to n+1
and n+1 to 1
each column adds up to n+2

so you have (n+1)(n+2) and we have to divide by 2 because we added it twice. (the two sums are multiplied as that is the function of multiplication - to ease addition)
simplifying (0.5)(n+1)(n+2):
=(0.5)(n+2)(n+1)
=(1/2n + 1)(n + 1)
which is correct

lol I concur.