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Sum of Product of Binomial Coefficients

If 1+x n = ...

Question

If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ , then the value of $C_{0} - C_{1} + C_{2} - C_{3} + . . . + {(- 1)}^{n} C_{n}$ is -`

A

2^{n}

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B

2^{n} - 1

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C

0

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D

`

2^{n - 1}

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Solution

The correct option is D $0$
$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . + C_{n} x^{x}$

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

mistake in question

$\Rightarrow$ right question
$C_{0} - C_{1} + C_{2} - C_{1} + . . . . . + (- 1)^{n} C_{n}$

Put $x = - 1$ in given ep.

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

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Similar questions

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

2 C_{0} + 2^{2} . \frac{C_{1}}{2} + 2^{3} . \frac{C_{2}}{3} + . . . + 2^{n + 1} . \frac{C_{n}}{n + 1} =

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

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . + C_{n - 2} 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}

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

C_{1} + C_{3} + C_{5} . . . . . . . . = ?

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C x^{2}

C_{0} + C_{1} + C_{2} + \dots + C_{n} = 2^{n}

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