To solve (a+b)[tex]^{n}[/tex], Pascal's triangle may help you:
1 n=0
1 1 n=1
1 2 1 n=2
1 3 3 1 n=3
1 4 6 4 1 n=4
1 5 10 10 5 1 n=5
1 6 15 20 15 6 1 n=6
1 7 21 35 35 21 7 1 n=7
1 8 28 56 70 56 28 8 1 n=8
I've shown the first 9 lines, however, you can continue writing lines indefinitely. Every number is the sum of upper two. For example in n=3 line, 1 = 0 + 1, 2 = 1 + 1, 1 = 1 + 0. In n=4 line 1 = 0 + 1, 3 = 1 + 2, 3 = 2 + 1, 1 = 1 + 0 and so on.
Numbers in these lines coefficients for expansion of (a+b)[tex]^{n}[/tex].
(a+b)[tex]^{n}[/tex] can be expanded to Ca[tex]^{n}[/tex]+Ca[tex]^{n-1}[/tex]b+Ca[tex]^{n-2}[/tex]b[tex]^{2}[/tex]+...+Ca[tex]^{2}[/tex]b[tex]^{n-2}[/tex]+Cab[tex]^{n-1}[/tex]+Cb[tex]^{n}[/tex] where C are the coefficients from line n.
For example:
(a+b)[tex]^{3}[/tex]=a[tex]^{3}[/tex]+3a[tex]^{2}[/tex]b+3ab[tex]^{2}[/tex]+b[tex]^{3}[/tex]
To expand (2+3x)[tex]^{5}[/tex] we must take the coefficients from n=5 line =>
(2+3x)[tex]^{5}[/tex]=2[tex]^{5}[/tex]+5*2[tex]^{4}[/tex]*3x+10*2[tex]^{3}[/tex]*(3x)[tex]^{2}[/tex]+10*2[tex]^{2}[/tex]*(3x)[tex]^{3}[/tex]+5*2*(3x)[tex]^{4}[/tex]+(3x)[tex]^{5}[/tex] = 32+270x+720x[tex]^{2}[/tex]+1080x[tex]^{3}[/tex]+810x[tex]^{4}[/tex]+243x[tex]^{5}[/tex]
If I calculated correctly, than from this you can see that the coefficient of x cubed is 1080. I think that expanding the binomial this way is much easier than multiplying.
I hope that helps.
P.S. If you see the triangle with a straight angle, than look http://en.wikipedia.org/wiki/Pascal_triangle" .