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Sum

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 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))`

Concept: Binomial Theorem for Positive Integral Indices

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