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Question

The coefficient of x7 in the expansion of (1xx2+x3)6(1x)12 is


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Solution

Expanding numerator and denominator and solving may not be feasible or might take so much time. We will to simplify it before expanding. In this case, we will try to find common factors numerator and denominator so that they will get cancelled.
(1xx2+x3)=[1x+(x2)(1x)]
=(1x)(1x2)
(1xx2+x3)6=((1x)(1x2))6
=(1x)(1x)(1+x))6
=(1x)12(1+x)6
(1xx2+x3)6(1x)12=(1+x)6
In the expansion of (1+x)6, there won't be any term with power 7
The coefficient of x7 is zero


Mathematics

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