$(x + 1)^{n} = C_{0} + C_{1} x^{1} + C_{2} x^{2} . . . . C_{n} x^{n}$
Find the value of $C_{0} - C_{1} + C_{2} - C_{3} . . . . C_{n}$ :

0

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2^{n - 1}

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2^{n}

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Cannot be determined.

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Solution

The correct option is A $0$
${(x + 1)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$
Putting $x = - 1$ and n is even.
$0 = C_{0} - C_{1} + C_{2} - . . . {- C}_{n - 1} + C_{n}$

If $x = - 1$ and n is odd,
$0 = C_{0} - C_{1} + C_{2} - . . . - C_{n}$
So, $A$ is correct.

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Similar questions

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

Find the value of

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Q. If

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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}$ , then $\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . . . . . + \frac{n C_{n}}{C_{n - 1}} =$

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(x+1)n=C0+C1x1+C2x2....CnxnFind the value of C0−C1+C2−C3....Cn :

The correct option is A 0(x+1)n=C0+C1x+C2x2+...+CnxnPutting x=−1 and n is even.0=C0−C1+C2−...−Cn−1+CnIf x=−1 and n is odd,0=C0−C1+C2−...−CnSo, A is correct.

$(x + 1)^{n} = C_{0} + C_{1} x^{1} + C_{2} x^{2} . . . . C_{n} x^{n}$
Find the value of $C_{0} - C_{1} + C_{2} - C_{3} . . . . C_{n}$ :

The correct option is A $0$
${(x + 1)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$
Putting $x = - 1$ and n is even.
$0 = C_{0} - C_{1} + C_{2} - . . . {- C}_{n - 1} + C_{n}$

If $x = - 1$ and n is odd,
$0 = C_{0} - C_{1} + C_{2} - . . . - C_{n}$
So, $A$ is correct.