The third term in the expansion of ${(2 x + \frac{1}{x^{2}})}^{m}$ does not contain $x$ . For which $x$ is that term equal to the second term in the expansion of ${(1 + x^{3})}^{30}$

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Solution

Given, ${(2 x + \frac{1}{x^{2}})}^{m}$
$T_{r + 1} =^{m} C_{r} x^{m - 3 r} 2^{m - r}$
For $r = 2,$ the term is independent of $x$ .
Hence
$m - 3 (2) = 0$
$m = 6$
$T_{3} =^{6} C_{2} 2^{4}$
$= 15 (16)$
$= 240$
$= T_{2}$ of $(1 + x^{3})^{30}$
Therefore
$30 x^{3} = 240$
$x^{3} = 8$
$x = 2$

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