The sum of all those terms which are rational numbers in the expansion of (2^1/3+3^1/4)^12…

Question

The sum of all those terms which are rational numbers in the expansion of $\left(2^{1/3}+3^{1/4}\right)^{12}$ is

Accepted Answer

$$T_{r+1}=\binom{12}{r}\left(2^{1/3}\right)^{12-r}\left(3^{1/4}\right)^r$$

Rational when $\dfrac{12-r}{3}$ and $\dfrac r4$ are both integers: $r\equiv0\pmod3$ and $r\equiv0\pmod4$, so $r\equiv0\pmod{12}$: $r=0$ or $r=12$.

- $r=0$: $T_1=\binom{12}{0}2^4=16$

Wait --- matching against source values: $r=0$ gives $T_1=3^0\cdot2^4=16$? Actually recompute: $(2^{1/3})^{12}=2^4=16$ at $r=0$, and $(3^{1/4})^{12}=3^3=27$ at $r=12$.

$T_1+T_{13}=16+27=43$

Solutions · 1