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Q.
Find the number of rational terms in the expansion of $\left(7^{1 / 3}+11^{1 / 9}\right)^{6561}$.
Binomial Theorem
Solution:
$T _{ r +1}={}^{6561} C _{ r }\left(7^{1 / 3}\right)^{6561- r } \cdot\left(11^{1 / 9}\right)^{ r }$
$={ }^{6561} C _{ r } 7^{2187} \cdot 7^{- r / 3} \cdot 11^{ r / 9}, 0 \leq r \leq 6561$
Now $T_{r}+1$ will be rational when $r / 3$ and $r/9$ both are integers.
This is possible only when $r$ is a multiple of $9$.
But multiple of $9$ from $0$ to $6561$ are
$0,9,18,27, \ldots \ldots \ldots, 6561$
These are in $AP$, so their number $n$ is given by
$6561=0+(n-1) 9$
$\Rightarrow n=730$