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Question

Prove that the coefficient of x5 in the expansion of (2x+3x2)6 is -4692

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Solution

(1+30x2x3)8
=n!(p!)(q!)(r!)(1)6(3x)q(2x3)r

where p + q + r = n = 8
for x5, we have q + 3r = 5
r = 0, q = 5 p = 3 by (I)
or r = 1, q = 2 p = 5 by (I)
Putting the above values in (A), we have
8!(3!)(5!)(0!).13.35(2)0+(8!)(5!)(2!)(1!).15.32.(2)1
=56(243)+8.7.6(9)=56(24354)=56(189)

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