The number of integral terms in the expansion | vedclass

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Question

${{\rm{T}}_{{\rm{r}} + 1}} = \sqrt[{650}]{{\rm{C}}}{\left( {{7^{\frac{1}{3}}}} \right)^{6551 - {\rm{r}}}}{\left( {{{11}^{\frac{1}{9}}}} \right)^x}$

Vedclass · Accepted Answer

$730$

Vedclass · Answer

${{m{T}}_{{m{r}} + 1}} = \sqrt[{650}]{{m{C}}}{\left( {{7^{\frac{1}{3}}}} ight)^{6551 - {m{r}}}}{\left( {{{11}^{\frac{1}{9}}}} ight)^x}$

$ = {\,^{6551}}{{m{C}}_r}{7^{2187 - \frac{r}{3}}}{11^{\frac{	au }{9}}}$

$ \Rightarrow \mathrm{r} =0,9,18, \ldots \ldots, 6561$

$\Rightarrow  \mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \Rightarrow 6561=0+(\mathrm{n}-1) 9$

$ \Rightarrow \mathrm{n} =730 $

The number of integral terms in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is :-

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