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7.Binomial Theorem
normal
The number of integral terms in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is :-
A
$721$
B
$730$
C
$745$
D
None of these
Solution
${{\rm{T}}_{{\rm{r}} + 1}} = \sqrt[{650}]{{\rm{C}}}{\left( {{7^{\frac{1}{3}}}} \right)^{6551 – {\rm{r}}}}{\left( {{{11}^{\frac{1}{9}}}} \right)^x}$
$ = {\,^{6551}}{{\rm{C}}_r}{7^{2187 – \frac{r}{3}}}{11^{\frac{\tau }{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 $
Standard 11
Mathematics
