I'm asked to find the coefficient of $x^3$ in $(2+x^2)^3 (3+2x)^7$.

For a simple problem like finding $x^2y^3$ in $(x+y)^5$ I can solve easily using binomial theorem.

But I have no idea how to go about approaching this one using binomial theorem, since I'm given two binomials(Each raised to separate exponents), multiplied by each other

Any help would be appreciated.

  • 2
    $\begingroup$ Why can't you use the binomial theorem here? You only get $x^3$ by multiplying a constant times a cube or an $x$ term by an $x^2$ term. You get each of those from each of the two factors using the binomial theorem and then you combine them. The first factor above doesn't have an odd powered term in $x$ so it is easier than the general case. $\endgroup$ – John Douma Nov 4 '15 at 22:27

This is about to get really REALLY tedious.

$ \sum\limits_{k=0}^{3} \binom{3}{k}(2)^{3-k}(x^2)^k \times \sum\limits_{m=0}^{7} \binom{7}{m}(3)^{7-m}(2x)^m $


$ \sum\limits_{k=0}^{3}\sum\limits_{m=0}^{7} \binom{3}{k}\binom{7}{m} (2)^{3-k + m} (3)^{7-m}(x)^{2k+m} $

Now you have just do this:

When 2k + m = 3

Case 1: k = 1, m = 1

Case 2: k = 0, m = 3

You plug that into your formula (remove the sigmas) and add up all the cases, so:

$ \binom{3}{1}\binom{7}{1} (2)^{3- 1 + 1} (3)^{7-1}(x)^{2+ 1} + \binom{3}{0}\binom{7}{3} (2)^{3- 0 + 3} (3)^{7-3}(x)^{2(0)+ 3} $

Good luck have fun :)

  • 1
    $\begingroup$ what about the case where $k=0$ and $m=3$? $\endgroup$ – John Joy Nov 4 '15 at 23:15
  • $\begingroup$ @JohnJoy Oh yea... oops. Thanks for catching that. $\endgroup$ – suchaHassle Nov 4 '15 at 23:16

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