Question

If n is a positive integer, find the coefficient of x^–1 in the expansion of `(1 + x)^2 (1 + 1/x)^n`

Answer 1

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 x^n–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))`