C 0 − C 1 + C 2 − C 3 + . . . . . . . . + ( − 1 ) n C n is equal to Options: A . &nbsp2n B . &nbsp2n−1 C . &nbsp0 D . &nbsp2n−1

C0−C1+C2−C3+........+(−1)nCn is equal to

Question

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

is equal to

Options:

A . 2n

B . 2n−1

C . 0

D . 2n−1

Answer: Option C
:
C
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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