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+x2−x)100=(1+x)(1+x)100(1+x2−x)100 =(1+x)((1+x)(1+x2−x))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