8/07/2014 Find the term in the expansion (3x – 2y)5 that contains x2y3.

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8/07/2014 Solution.
Since, in general, the kth term of (x + y)n is nCk – 1 x (nk)·yk, that for (3x – 2y)5 is
5Ck1 (3x) (5 – k)·(-2y)k so that with k = 3 – because the exponent of y should be 3 – we have 5C31 (3x) (5 – 3)·(-2y)3 = -720x2y3.