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Question

Number of terms in the expansion of (1+x)101(1+x2−x)100 is

A
302
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B
301
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C
202
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D
101
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Solution

The correct option is C 202
(1+x)101(1+x2x)100=(1+x)(1+x)100(1+x2x)100
=(1+x)((1+x)(1+x2x))100
=(1+x)(1+x3)100
There are two types of terms here:
x3k type and x(3k+1) type
So, number of terms are 101+101=202 terms

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