Question

If $p$ is an integral multiple of $4$ lying in between the coefficients of $x^4$ and $x$ in the expansion of ${\left(x^2+\frac{1}{x}\right)}^8$, then the number of such values of $p$ is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (A)

The general term in the expansion of ${\left(x^2+\frac{1}{x}\right)}^8$ is
$T_{r+1}={}^8{C}_r{\left(x^2\right)}^{8-r}{\left(\frac{1}{x}\right)}^r={}^8{C}_r{x}^{16-3r}$
For the coefficient of $x^4$, put $r=4$, we get
${}^8{C}_4=\frac{8\times 7\times 6\times 5}{4\times 3\times 2}=70$
For the coefficient of $x$, put $r=5$, we get
${}^8{C}_5=\frac{8\times 7\times 6}{3\times 2}=56$
Now, the numbers which are integral multiple of $4$ and lying in between $56$ and $70$ are $60,64$ and $68$ .