JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q67

MathematicsSingle correct+4 / -1Question 67 of 90

The number of terms in the expansion of $(1+x)^{101}\left(1+x^2-x\right)^{100}$ in powers of $x$ is :

302

301

202

101

Answer and worked solution

Official answer: C

$(1+x)^{101}(1-x+x^2)^{100}=(1+x)\left[(1+x)(1-x+x^2)\right]^{100}=(1+x)(1+x^3)^{100}$ .

This has terms $x^{3k}$ and $x^{3k+1}$ for $k=0,\dots,100$ , i.e. $2\times101=202$ terms.

JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 67