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`(1)/("log"_(2)n) + (1)/("log"_(3)n) + (1)/("log"_(4)n) + … + (1)/("log"_(43)n)=`A. 1B. `"log"_(43!)n`C. `"log"_(n)43!`D. none of these |
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Answer» Correct Answer - C We have, `(1)/("log"_(2)n) + (1)/("log"_(3)n) + (1)/("log"_(4)n) + … + (1)/("log"_(43)n)` ` = "log"_(n)2 + "log"_(n)3 + "log"_(n)4 +… +"log"_(n)43` ` = "log"_(n) (2 xx 3 xx 4 xx .. xx 43) = "log"_(n)43!` |
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