Question

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

• A. $84$
• B. $75$
• C. $68$
• D. $36$

Answer 1

Answer: (A)

$84$

The first three coefficients will always be

${}^m{C}_0=1$

${}^m{C}_1=m$

${}^m{C}_2=\frac{m(m-1)}{2}$

The sum of these simplifies to

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

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

$\frac{m^2}{2}-\frac{m}{2}+m+1=46$

$\frac{m^2}{2}+\frac{m}{2}+1=46$

$\Rightarrow m^2+m+2=92$

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

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

The only positive solution is $m=9$

Now in the expansion with $m=9$ the term lacking $x$ must be the term containing

${\left(x^2\right)}^3{\left(\frac{1}{x}\right)}^6=\frac{x^6}{x^6}=1$

This term has a coefficient of ${}^9{C}_6=\frac{9!}{3!6!}=\frac{9\times 8\times 7\times 6!}{3!6!}=\frac{9\times 8\times 7}{3\times 2\times 1}=84$

The solution is $84$