The binomial expansion tells us how to write the expression (a+b)N in terms of a sum of various powers of a and b. The way to do this in principle is just to write out all the factors and expand term by term. For example
(a+b)2 = (a+b) x (a+b) = a x (a+b) + b x (a+b)
= aa + ab + ba + bb
= a2 + ab + ab +b2 (note: this expression has 4 terms)
= a2 + 2ab + b2
where ab = a x b, and from now on we will stop using the "x" symbol for multiplication, and just indicate multiplication by writing the two quantities next to each other.
(a+b)3 = (a+b)(a+b)(a+b) = a(a+b)(a+b) + b(a+b)(a+b)
= aa(a+b) + ab(a+b) + ba(a+b) + bb(a+b)
= aaa + aab + aba + abb + baa + bab + bba + bbb
= a3 + a2b + a2b + ab2 + a2b + ab2 + ab2 + b3 (note: the above expression has 8 terms)
= a3 + 3a2b + 3ab2 + b3
(a+b)N = (a+b)(a+b)(a+b)...(a+b)
where "..." means there are a total of N such terms. When we expand out, each resulting term can be viewed as if it were the result of "choosing" either an "a" or a "b" from each of the N factors (a+b). Since there are two such choices for each factor (a+b), we see that we will wind up with 2N terms (check that this agrees with the above cases: for N=2 we found 22=4 terms, for N=3 we found 23=8 terms).
Consider now the particular case where we have chosen "b" r times and hence "a" is chosen the remaining N-r times, we wind up with the term aN-rbr. There will be many such terms aN-rbr for a given value of r, as there are many different ways to choose the r "b"s from the N factors (a+b). How many ways are there? There are exactly NCr ways! So when we expand (a+b)N, the term aN-rbr appears times. We thus can group them all together into a single term:
Now there will such terms for all values of r = 0, 1, 2, ..., N. Note that for the term r=0 we choose "b" no times, and choose "a" N times. The resulting factor, aN, is consistent with the above formula provided we regard 0!=1, as we said earlier in our definition of the factorial, and also b0=1. This last result b0=1 is consistent with the usual mathematical conventions of power laws. Similarly, for the term r=N, the above formula is consistent with bN.
So finally we have the binomial expansion:
Where we use the fact that x0
= 1, and x1 = x, for any number x. You
can check that this formula gives the correct results for the cases N=2
and N=3 which we did explicitly above.