In the expansion of $(1 - x)^{n}$ , the sum of the odd terms is $S_{1}$ , and the sum of the even terms is $S_{2}$ , then $S_{1} - S_{2} =$

0

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(1 + x)^{n}

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- (1 - x)^{n}

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(1 - x)^{2 n}

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Solution

The correct option is B $(1 + x)^{n}$
$(1 - x)^{n}$
For the above the sum of all odd terms will be ( $S_{1}$ )
$= C_{0} + C_{2} x^{2} . . .$
And sum of all even terms will be $(S_{2})$
$= - C_{1} x - C_{3} x^{3} . . .$
Therefore $S_{1} - S_{2}$ will be
$C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} . . . C_{n} x^{n}$
$= (1 + x)^{n}$
Hence answer is Option B

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In the expansion of (1−x)n, the sum of the odd terms is S1, and the sum of the even terms is S2, then S1−S2=

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In the expansion of $(1 - x)^{n}$ , the sum of the odd terms is $S_{1}$ , and the sum of the even terms is $S_{2}$ , then $S_{1} - S_{2} =$