(1+x)101(1+x2−x)100
=(1+x)(1+x)100(1−x+x2)100
=(1+x)(1+x3)100 ∵a3+b3=(a+b)(a2−ab+b2)
(1+x3)100 has 101 terms
(1+x)(1+x3)100=1(1+x3)100+x(1+x3)100
⇒ No. of terms = 101+101=202
The number of terms in the expansion of (1+x)101(1−x+x2)100 in powers of x is
Held on 9 Apr 2014 · Verified 6 Jul 2026.
301
302
101
202
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