The sum of all rational terms in the expansio | vedclass

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(B) The general term in the expansion of $(1+2^{1/3}+3^{1/2})^6$ is given by $\frac{6!}{r_1! r_2! r_3!} (1)^{r_1} (2^{1/3})^{r_2} (3^{1/2})^{r_3}$,whe

Vedclass · Accepted Answer

$612$

Vedclass · Answer

(B) The general term in the expansion of $(1+2^{1/3}+3^{1/2})^6$ is given by $\frac{6!}{r_1! r_2! r_3!} (1)^{r_1} (2^{1/3})^{r_2} (3^{1/2})^{r_3}$,where $r_1+r_2+r_3=6$.For the term to be rational,$r_2$ must be a multiple of $3$ and $r_3$ must be a multiple of $2$.Possible values for $(r_1, r_2, r_3)$ are:$1$. $(6, 0, 0) \implies \frac{6!}{6!0!0!} (1)^6 (2)^0 (3)^0 = 1$$2$. $(4, 0, 2) \implies \frac{6!}{4!0!2!} (1)^4 (2)^0 (3)^1 = 15 	imes 3 = 45$$3$. $(2, 0, 4) \implies \frac{6!}{2!0!4!} (1)^2 (2)^0 (3)^2 = 15 	imes 9 = 135$$4$. $(0, 0, 6) \implies \frac{6!}{0!0!6!} (1)^0 (2)^0 (3)^3 = 1 	imes 27 = 27$$5$. $(3, 3, 0) \implies \frac{6!}{3!3!0!} (1)^3 (2)^1 (3)^0 = 20 	imes 2 = 40$$6$. $(1, 3, 2) \implies \frac{6!}{1!3!2!} (1)^1 (2)^1 (3)^1 = 60 	imes 6 = 360$$7$. $(0, 6, 0) \implies \frac{6!}{0!6!0!} (1)^0 (2)^2 (3)^0 = 1 	imes 4 = 4$Sum $= 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612$.

The sum of all rational terms in the expansion of $(1+2^{1/3}+3^{1/2})^6$ is equal to . . . . . . .

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