The sum of the rational terms in the binomial | vedclass

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

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(D) The general term in the expansion of $(2^{1/2} + 3^{1/5})^{10}$ is given by $T_{r+1} = ^{10}C_r (2^{1/2})^{10-r} (3^{1/5})^r = ^{10}C_r (2)^{(10-r)/2} (3)^{r/5}$.For the term to be rational,the exponents of $2$ and $3$ must be integers.Thus,$(10-r)/2$ must be an integer,which implies $r$ must be even.Also,$r/5$ must be an integer,which implies $r$ must be a multiple of $5$.Since $0 \le r \le 10$,the possible values for $r$ are $0$ and $10$.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 $(2^{1/2} + 3^{1/5})^{10}$ is

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