Question

The coefficient of $x^{65}$ in the expansion of $(1+x{)}^{131}(x^2-x+1{)}^{130}$ is?

• A. ${}^{130}{C}_{65}+$ ${}^{129}{C}_{66}$
• B. ${}^{130}{C}_{65}+$ ${}^{129}{C}_{55}$
• C. ${}^{130}{C}_{.66}+$ ${}^{129}{C}_{65}$
• D. $Noneofthese$

Answer 1

Answer: (D)

$Noneofthese$

${(1+x)}^{131}{(x^2-x+1)}^{130}=(1+x){\left((1+x)(x^2-x+1)\right)}^{130}$

$=(1+x){(1+x^3)}^{130}={(1+x^3)}^{130}+x{(1+x^3)}^{130}$

$x^{65}$ coefficient in ${(1+x^3)}^{130}$

${(1+x^3)}^{130}={{}^{130}C}_0{\left(x^3\right)}^0+{{}^{130}C}_1{\left(x^3\right)}^1+.....+{{}^{130}C}_{130}{\left(x^3\right)}^{130}={\sum }_{r=0}^{\infty }{{}^{130}C}_0{x}^{3r}$

$3r=65\Rightarrow r=\frac{65}{3}$ but $r$ is +ve integer

So, there is no $x^{65}$ term in ${(1+x^3)}^{130}$

for $x^{65}$ coefficient in ${(1+x^3)}^{130}$, we need to find $x^{64}$ in ${(1+x^3)}^{130}$

$3r=64\Rightarrow r$ is not an integer

The coefficient of $x^{65}=0$