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

2^{n}

B .

2^{n} - 1

C . 0

D .

2^{n - 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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