Which of the following is the greatest?

Question

(A)  \frac{1}{√2}+\frac{1}{√4}+\frac{1}{√6}+\frac{1}{√8}

(B)  \frac{1}{{2^2}}+\frac{1}{{4^2}}+\frac{1}{{6^2}}+\frac{1}{{8^2}}

(C)  \frac{1}{{2^2}}+\frac{1}{{2^4}}+\frac{1}{{2^6}}+\frac{1}{{2^8}}

(D)  1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}

(E)  \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}

Abhishekgmat87 · Answer

Sum of 1st 2 terms in option A is greater than 1. In option B,C and D, the denominators are very large. They will never add up to 1. In option E, the sum is less tha. 1

Answer is Option A

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RositaDhe · Answer

Lemme try to explain as a beginner.

A). 1/√2+1/√4+1/√6+1/√8 = 1/2^ 1/2 , .... , 1/8^ 1/2

B). 1/2^ 2 ,..., 1/8^ 2

In contrast, B). C). D). are Larger, and Larger. The Fraction rules...'Smallest denominator, Greatest Value'

KarishmaB · Answer

Compare pairs of options and eliminate.

First (D) and (E) since they are pretty simple.

Since 1 - 1/2 = 1/2, (D) becomes

(D). \frac{1}{2}+\frac{1}{4}-\frac{1}{6} 
 (E). \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}

(E) is greater. Eliminate (D).

(C). \frac{1}{{2^2}}+\frac{1}{{2^4}}+\frac{1}{{2^6}}+\frac{1}{{2^8}} 
 (E). \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}

Every denominator of (C) is greater than every corresponding denominator of (E). So (C) has smaller terms. Eliminate (C). Same logic for (B).

(A). \frac{1}{√2}+\frac{1}{√4}+\frac{1}{√6}+\frac{1}{√8} 
 (E). \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}

√2 < 2, √4 < 4, √6 < 6 and √8 < 8
Hence all denominators of option (A) are smaller and hence, all terms of option (A) are greater than corresponding terms of option (E).

Hence option (A) is the greatest.

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Which of the following is the greatest?

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Which of the following is the greatest?

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