The absolute value of middle term in the expansion of 1-1-x n-1-xn is

The correct option is D ^2nC_n Given, (1 1x)^n(1 x)^n =1x^n(x 1)^n(1 x)^n =( 1)^nx^n[1 x]^2n Therefore the middle term will be T_n+1= ^ ...

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First Principle of Differentiation

The absolute ...

Question

The absolute value of middle term in the expansion of ${(1 - \frac{1}{x})}^{n} . (1 - x)^{n}$ is

^{2 n} C_{n}

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-^{2 n} C_{n}

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-^{2 n} C_{n - 1}

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none of these

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(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

\sum \sum 0 \leq r < s \leq n {(C_{r} + C_{s})}_{r}^{2}

The middle term in the expansion of $(x + \frac{1}{x})^{2 n}$ is:

The middle term in the expansion of $(x^{2} + \frac{1}{x^{2}} + 2)^{n}$ is:

^{2 n} C_{n} = C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + \dots \dots + C_{n}^{2}

^{2 n} C_{n} =

x

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

^{2 n} C_{n} = \frac{1.3.5.7 \dots \dots (2 n - 1)}{n!}

(1 + x)^{n} = C_{0} + C_{1} x + \dots . . + C_{n} x^{n}

\sum \sum 0 \leq r < s \leq n C_{r} C_{s}

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