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Question

Let a and b be the coefficient of x3 in (1+x+2x2+3x3)4 and (1+x+2x2+3x3+4x4)4 respectively. Then the value of 6ab is

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Solution

b=coefficient of x3 in ((1+x+2x2+3x3)+4x4)4
=coefficient of x3 in [4C0(1+x+2x2+3x3)4(4x4)0 + 4C1(1+x+2x2+3x3)3(4x4)1+...]

Neglect the higher power term, we get
=coefficient of x3 in (1+x+2x2+3x3)4
=a
Hence, 6ab=6

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