Question

If the sum of the coefficients of the first, second and third terms of the expansion of ${(x^2+\frac{1}{x})}^m$ is $46$, then the coefficient of the term that does not contain $x$ is

• A. $84$
• B. $92$
• C. $98$
• D. $106$

Answer 1

The correct option is A $84$

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