The Binomial Theorem says that, for a positive integer n,
(x+b)^n= ""_nC_0 x^n + ""_nC_1 x^(n-1)b+""_nC_2 x^(n-2)b^2 + ... + ""_nC_n b^n
This can be succinctly written as the sum
(x+b)^n= sum_(k=0)^n ""_nC_k x^(n-k)b^k
To see why this formula works, let's use it on the binomial for this question, (x+2)^4. If we were to write out all the factors side-by-side, we'd get
(x+2)^4=(x+2)(x+2)(x+2)(x+2)
Multiplying this out means making all possible products of one term from each binomial, and adding these products together. Sort of like FOIL-ing to the next level. (In FOIL-ing, there are color(blue)2 binomials, so there will be 2^color(blue)2=4 terms; with color(green)4 binomials, there will be 2^color(green)4=16 terms.)
After (tediously) multiplying it all out, we'll get some terms with x^4, some with x^3 * 2, some with x^2 * 2^2, and so on. But how many of each?
Since there is only ""_4C_0=1 way to pair up all four x's, we'll only get a single x^4 term.
There are ""_4C_1=4 ways to make an x^3 * 2 term—using the 2 from binomials 1, 2, 3, and 4 (multiplied with the x's from the others).
Continuing this way, there will always be ""_4C_k ways to make an x^(n-k) * 2^k term. So this is where the combination notation in the Binomial Theorem comes from. For k=0,...,4, the x^(n-k) * 2^k bit (from the sum above) represents the different terms we can make, and the ""_4C_k bit is the number of ways to make that term.
Thus, the binomial expansion of (x+2)^4 is
""_4C_0 x^4 + ""_4C_1 x^3 * 2 + ""_4C_2 x^2 * 2^2 + ""_4C_3 x * 2^3 + ""_4C_4 * 2^4
=1x^4+4x^3(2)+6x^2(4)+4x(8)+1(16)
=x^4+8x^3+24x^2+32x+16.
