The number of positive integers k such that t | vedclass

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Question

$\left(2 x^{3}+\frac{3}{x^{k}}\right)^{12}$

$t _{ r +1}={ }^{12} C _{ r }\left(2 x ^{3}\right)^{ r }\left(\frac{3}{ x ^{ k }}\right)^{12- r }$

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Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\left(2 x^{3}+\frac{3}{x^{k}}ight)^{12}$

$t _{ r +1}={ }^{12} C _{ r }\left(2 x ^{3}ight)^{ r }\left(\frac{3}{ x ^{ k }}ight)^{12- r }$

$x ^{3 r -(12- r ) k } ightarrow constant$

$	herefore 3 r -12 k + rk =0$

$\Rightarrow k =\frac{3 r }{12- r }$

$	herefore$ possible values of $r$ are $3,6,8,9,10$ and corresponding values of $k$ are $1,3,6,9,15$

Now ${ }^{12} C _{ r }=220,924,495,220,66$

$	herefore$ possible values of $k$ for which we will get $2^{8}$ are $3. 6$

The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2 x^{3}+\frac{3}{x^{k}}\right)^{12}, x \neq 0$ is $2^{8} \cdot \ell$, where $\ell$ is an odd integer, is......

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