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Question
If n is a positive integer, find the coefficient of x–1 in the expansion of `(1 + x)^2 (1 + 1/x)^n`
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Solution
We have `(1 + x)^n 1 + 1^n/x`
= `(1 + x)^n (x + 1)^n/x`
= `(1 + x)^(2n)/x^n`
Now to find the coefficient of x–1 in `(1 + x)^n 1 + 1^n/x`
It is equivalent to finding coefficient of x–1 in `(1 + x)^(2n)/x^n`
Which in turn is equal to the coefficient of xn–1 in the expansion of (1 + x)2n
Since (1 + x)2n = `""^(2n)"C"_0 x^0 + ""^(2n)"C"_1 + x^1 + ""^(2n)"C"_2 x^2 + ... + ""^(2n)"C"_(n - 1) x^(n - 1) + ... + ""^(2n)"C"_(2n) x^(2n)`
Thus the coefficient of `x^(n - 1)` is `""^(2n)"C"_(n - 1)`
= `(2n)/((n - 1)(2n - n + 1))`
= `(2n)/((n - 1)(n + 1))`
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