The number of irrational terms in the expansi | vedclass

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(D) Let $E = (5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$.Using the binomial expansion,the general term $T_{r+1}$ of $(a+b)^n + (a-b)^n$ i

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$384$

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(D) Let $E = (5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$.Using the binomial expansion,the general term $T_{r+1}$ of $(a+b)^n + (a-b)^n$ is $2 	imes \sum_{k=0, 2, 4, \dots} \binom{n}{k} a^{n-k} b^k$.Here $n = 1024$,$a = 5^{1/2}$,and $b = 7^{1/8}$.The terms are of the form $2 \binom{1024}{k} (5^{1/2})^{1024-k} (7^{1/8})^k = 2 \binom{1024}{k} 5^{512 - k/2} 7^{k/8}$.For the term to be rational,$k/2$ and $k/8$ must be integers,which implies $k$ must be a multiple of $8$.Since $0 \le k \le 1024$ and $k$ is even,$k \in \{0, 8, 16, \dots, 1024\}$.The number of such values of $k$ is $\frac{1024}{8} + 1 = 128 + 1 = 129$.These $129$ terms are rational.The total number of terms in the expansion of $(5^{1/2} + 7^{1/8})^{1024}$ is $1024 + 1 = 1025$.However,in the sum $(5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$,the odd terms cancel out,leaving only the even terms.The number of terms in the resulting expansion is $\frac{1024}{2} + 1 = 513$.Out of these $513$ terms,$129$ are rational.Therefore,the number of irrational terms is $513 - 129 = 384$.

The number of irrational terms in the expansion of $(5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$ is

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