$^{14} C_{1} +^{14} C_{2} +^{14} C_{3} + . . . . +^{14} C_{14} =$

2^{14}

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2^{14} - 1

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2^{14} + 2

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2^{14} - 2

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Solution

The correct option is B $2^{14} - 1$
Consider expansion of $(1 + x)^{14}$
$(1 + x)^{14} = \sum_{k = 0}^{14} (^{14} C_{k} x^{k})$
Substitute x=1 in above expression.
$(1 + 1)^{14} =^{14} C_{0} +^{14} C_{1} + . . . . . .^{14} C_{14}$
$\Rightarrow^{14} C_{1} + . . . . . .^{14} C_{14} = 2^{14} -^{14} C_{0} = 2^{14} - 1$ .

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