The term independent of x in 1-xm1-1xn is-

The correct option is B ^m+nC_n A](1+x)^m(x+1/x)^n =(1+x/x^n)^m+n expanding, ^m+nC_ox^o+...+^m+nC_m+nC_x^m+n/x^n only ^m+nC_nx^n is a independ ...

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Combination

The term inde...

Question

The term independent of $x$ in $(1 + x)^{m} {(1 + \frac{1}{x})}^{n}$ is:

^{m + n} C_{m}

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^{m + n} C_{n}

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^{m + n} C_{m - n}

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N o n e o f t h e s e

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m

n

^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + . . . . +^{m} C_{k}^{n} C_{0}

(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{n} x^{n}, x ϵ N

^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - . . . . . . + (- 1)^{m - 1} \times^{n} C_{m - 1}

= (- 1)^{m - 1} \frac{(n - 1) (n - 2) . . (n - m + 1)}{(m - 1)!}

(w i t h n \geq m)

(\begin{matrix} n m \end{matrix}) =^{n} C_{m}

(\begin{matrix} n m \end{matrix}) + (\begin{matrix} n - 1 m \end{matrix}) + (\begin{matrix} n - 2 m \end{matrix}) + . . . . . . . . . . + (\begin{matrix} m m \end{matrix}) = (\begin{matrix} n + 1 m + 1 \end{matrix})

m, n (with n \geq m) =^{n} C_{m}

(\begin{matrix} n m \end{matrix}) + (\begin{matrix} n - 1 m \end{matrix}) + (\begin{matrix} n - 2 m \end{matrix}) + . . . + (\begin{matrix} m m \end{matrix}) =

x

(1 + x)^{n} . {(1 - \frac{1}{x})}^{n}

(n \in N)

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