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Question

Number of term in (1+x)101(1+x2x)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 A 202
f(x)=(1+x)101(1+x2x)100

=(1+x)[(1+x)100(x2x+1)]100

=(1+x)(1+x3)100

Now number of terms in the expansion of (a+b)n=n+1

Hence number of terms in the expansion of (1+x3)100=101
So number of term in the expansion of

f(x)=(x+1)(x3+1)100=x[(x3+1)100]+[(x3+1)100]

=101+101=2×101=202

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