Question

The sum of the coefficients in the first, second, and third terms of the expansion of ${(x^2+\frac{1}{x})}^m$ is equal to $46$. Find the term of the expansion which does not contain $x$.

Answer 1

coeff of first term $={}^m{C}_0$

coeff of second term $={}^m{C}_1$

coeff of third term $={}^m{C}_2$

Given ,${}^m{C}_0+{}^m{C}_1+{}^m{C}_2=46$

$\frac{m!}{0!m!}+\frac{m!}{(m-1)!1!}+\frac{m!}{(m-2)!2!}=46$

$\Rightarrow m+\frac{(m-1)m}{2}=45$

$\Rightarrow 2m+(m-1)m=90$

$\Rightarrow m^2+m-90=0$

$\Rightarrow m^2+10m-9m-90=0$

$\Rightarrow (m+10)(m-9)=0$

$\therefore m=9$ [since $m$ cannot have negative value]

$T_{r+1}={}^9{C}_r{\left(x^2\right)}^{9-r}{\left(\frac{1}{x}\right)}^r$

$T_{r+1}={}^9{C}_r{x}^{18-2r-r}={}^9{C}_r{x}^{18-3r}$

Therefore term of the expansion which does not contain $x$ is

$18-3r=0$

$\Rightarrow r=6$

$\therefore T_{6+1}=T_7$

i.e, $7^{th}$ term does not contain x