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Q. Sum of all the rational terms in the expansion of $\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12},$ is

Binomial Theorem

Solution:

[$T_{r+1}={ }^{12} C_{r}\left(3^{\frac{1}{4}}\right)^{12-r} \cdot(4)^{r / 3}$
$={ }^{12} C_{r} \cdot 3^{(3-r / 4)} \cdot(4)^{r / 3}$
For rational term $r=0$ or $12$
$\therefore $ sum $=T_{1}+T_{13}={ }^{12} C_{0} \cdot 3^{3}+{ }^{12} C_{12} \cdot 4^{4} $
$=3^{3}+4^{4}=27+256=283$