# If n is a positive integer, find the coefficient of x–1 in the expansion of (1+x)2(1+1x)n - Mathematics

Sum

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

#### 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))

Concept: Binomial Theorem for Positive Integral Indices
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Solved Examples | Q 12 | Page 136
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