Question

The number of terms in the expansion off $(1+x{)}^{101}(1+x^2-x{)}^{100}$ in power of x is :

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

Answer 1

The correct option is C $202$

${(1+x)}^{101}{(1-x+x^2)}^{100}$ can be written as

$=(1+x){(1+x)}^{100}{(1-x+x^2)}^{100}$

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

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

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

Number of terms in ${(1+x^3)}^{100}=101$[terms are $a_0+a_1{x}^3+a_2{x}^6+............+a_{1}00x^{300}$]

Number of terms in $x{(1+x^3)}^{100}=101$[terms are $a_{0}x+a_1{x}^4+a_2{x}^7+............+a_{1}00x^{301}$]

since there are no common terms

.
So, number of terms in $(1+x){(1+x^3)}^{100}=101+101=202$