C 0- C 1- C 2- C 3-1 n C n is equal toA- 2nB- 2-1C- 2n-1D- 0

The correct option is D 0 We know that (1 + x)^n = nbsp;^n C nbsp;_0 nbsp;+ nbsp; nbsp;^n C nbsp;_1 nbsp;x + nbsp; nbsp;^n C nbsp;_2 nbs ...

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Multinomial Expansion

C 0- C 1+ C 2...

Question

$C_{0} - C_{1} + C_{2} - C_{3} + . . . . . . . . + (- 1)^{n} C_{n}$ is equal to

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$C_{0} - C_{1} + C_{2} - C_{3} + . . . . . . . . + (- 1)^{n} C_{n}$ is equal to

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . + C_{n} x^{n}

(C_{0} - C_{1} + C_{2} - C_{3} + . . . . . + (- 1)^{n} C_{n})

C_{0}, C_{1}, C_{2}, . . . . . . . . . . . C_{n}

(1 + x)^{n} .

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . . (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1})

C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \frac{C_{3}}{4} + . . . + {(- 1)}^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1}

\frac{C_{0}}{1.3} - \frac{C_{1}}{2.3} + \frac{C_{2}}{3.3} - \frac{C_{3}}{4.3} + . . . + {(- 1)}^{n} \frac{C_{n}}{(n + 1) .3}

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