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
$T_{r+1}={ }^{8} C_{r}\left(x^{2}\right)^{8-r}\left(\frac{1}{x}\right)^{r}={ }^{8} C_{r} x^{16-3 r}$
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 $68$ .