Prove that $C_{0} + C_{1} + C_{2} + . . . . . C_{n} = 2^{n}$

2^{n}

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2^{n - 1}

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2^{n + 1}

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None of the above.

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Solution

The correct option is A $2^{n}$
$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x + \dots +^{n} C_{n} x^{n}$

Let $x = 1$

$\Rightarrow C_{0} + C_{1} .1 + C_{2} {.1}^{2} + \dots + C_{n} {.1}^{n} = (1 + 1)^{n} = 2^{n}$

$\Rightarrow C_{0} + C_{1} + C_{2} .1 + \dots + C_{n} = (1 + 1)^{n} = 2^{n}$

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