Question

If sum of the coefficients of first, second and third terms in the expansion of ${(x^2+\frac{1}{x})}^m$ is $46,$ then the coefficient of the term that is independent of $x,$ is

• A. $96$
• B. $84$
• C. $78$
• D. $88$

Answer 1

The correct option is B $84$

We are given
${}^m{C}_0+{}^m{C}_1+{}^m{C}_2=46$
$\Rightarrow 1+m+\frac{m(m-1)}{2}=46$
$\Rightarrow 2m+m(m-1)=90$
$\Rightarrow m^2+m-90=0$
$\Rightarrow m=9$ or $m=-10$
$\Rightarrow m=9$as $m>0$
Now, $(r+1)th$ term of ${(x^2+\frac{1}{x})}^m$ is ${}^m{C}_r(x^2{)}^{m-r}{\left(\frac{1}{x}\right)}^r$
$={}^m{C}_r{x}^{2m-3r}$
For this to be independent of $x$, $2m-3r=0\Rightarrow r=6$
$\therefore$ Coefficient of the term independent of $x$ is ${}^9{C}_6=84.$