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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
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The correct answer is:
$3$
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 . Hence, option (a) is correct.
$\left(x^2+\frac{1}{x}\right)^8$ 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 . Hence, option (a) is correct.
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