The sum of the rational terms in the binomial | vedclass

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(A) The general term in the expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is given by $T_{r+1} = {}^{10}C_r (\sqrt{2})^{10-r} (3^{1/5})^r = {}^{10}C_r (2)^

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(A) The general term in the expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is given by $T_{r+1} = {}^{10}C_r (\sqrt{2})^{10-r} (3^{1/5})^r = {}^{10}C_r (2)^{(10-r)/2} (3)^{r/5}$.For the term to be rational,the exponents of $2$ and $3$ must be integers.Thus,$(10-r)/2$ must be an integer (which is true for all even $r$) and $r/5$ must be an integer.For $0 \le r \le 10$,the values of $r$ that satisfy both conditions are $r = 0$ and $r = 10$.For $r = 0$: $T_1 = {}^{10}C_0 (2)^5 (3)^0 = 1 	imes 32 	imes 1 = 32$.For $r = 10$: $T_{11} = {}^{10}C_{10} (2)^0 (3)^2 = 1 	imes 1 	imes 9 = 9$.The sum of the rational terms is $32 + 9 = 41$.

The sum of the rational terms in the binomial expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is

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