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

$2^{n}$

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$2^{n} - 1$

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$2^{n - 1}$

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Solution

The correct option is C
0

We know that

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

Putting x = -1, we get

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

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

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Similar questions

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}

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

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

C_{n}

(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})

\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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