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If `n=1999!` then `sum_(x=1)^(1999) log_n x=`A. 1B. 0C. `root(1999)(1999)`D. `-1` |
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Answer» Correct Answer - A We have, `sum_(x =1)^(1999) "log"_(n) x` ` ="log"_(n) 1 + "log"_(n)2 + "log"_(n) 3 +. … + "log"_(n) 1999` `="log"_(n) (1*2*3……..1999) = "log"_(n) 1999! = "log"_(n) n = 1` |
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