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The term independent of x in the expansion of(1+x)^n\left(1+\frac{1}{x}\right)^n  is

Option: 1

C_0^2+2 C_1^2+\ldots \ldots+(n+1) C_n^2


Option: 2

\left(C_0+C_1+\ldots \ldots+C_n\right)^2


Option: 3

C_0^2+C_1^2+\ldots \ldots+C_n^2


Option: 4

None of these


Answers (1)

best_answer

We know that, (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots \ldots \ldots .+{ }^n C_n x^n

\left(1+\frac{1}{x}\right)^n={ }^n C_0+{ }^n C_1 \frac{1}{x^1}+{ }^n C_2 \frac{1}{x^2}+\ldots . .+{ }^n C_n \frac{1}{x^n}

Obviously, the term independent of x will be

{ }^n C_0 \cdot{ }^n C_0+{ }^n C_1{ }^n C_1+\ldots . .+{ }^n C_n \cdot{ }^n C_n=C_0^2+C_1^2+\ldots \ldots+C_n^2

Trick : Put n=1 in the expansion of  (1+x)^1\left(1+\frac{1}{x}\right)^1=1+x+\frac{1}{x}+1=2+x+\frac{1}{x}\,...(i)

We want coefficient of x^0. Comparing to equation (i). Then, we get 2 i.e., independent of x.

Option (c): C_0^2+C_1^2+\ldots . . C_n^2; Put\,\, n=1; \,Then\,\, { }^1 C_0^2+{ }^1 C_1^2=1+1=2

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shivangi.bhatnagar

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