Question

Number of term in ${(1+x)}^{101}{(1+x^2-x)}^{100}$ is

• A. $302$
• B. $301$
• C. $202$
• D. $101$

Answer 1

Answer: (C)

$202$

$f\left(x\right)={(1+x)}^{101}{(1+x^2-x)}^{100}$

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

$=(1+x)(1+x^3{)}^{100}$

Now number of terms in the expansion of $(a+b{)}^n=n+1$

Hence number of terms in the expansion of $(1+x^3{)}^{100}=101$

So number of term in the expansion of

$f\left(x\right)=(x+1)(x^3+1{)}^{100}=x[(x^3+1{)}^{100}]+\left[\right(x^3+1{)}^{100}]$

$=101+101=2\times 101=202$