If n-2006- then 1-log2n-1-log3n-1-log2006n-

Explanation for correct option Given n=2006!∴1/log_2(n)+1/log_3(n)+....+1/log_2006(n)0ex0ex=log_n(2)+log_n(3)+..+log_n(2006) [∵1/log_a(b)=log_b(a)]0ex0e ...

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Standard XII

Mathematics

Integration Using Substitution

If n=2006! th...

Question

If $n = 2006!$ then $\frac{1}{\log_{2} (n)} + \frac{1}{\log_{3} (n)} + . . . . + \frac{1}{\log_{2006} (n)} =$

$2006$

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$2005$

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$2005!$

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$1$

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$0$

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Solution

The correct option is D
$1$

Explanation for correct option
Given $n = 2006!$
$∴ \frac{1}{\log_{2} (n)} + \frac{1}{\log_{3} (n)} + . . . . + \frac{1}{\log_{2006} (n)} = \log_{n} (2) + \log_{n} (3) + . . + \log_{n} (2006) [∵ \frac{1}{\log_{a} (b)} = \log_{b} (a)] = \log_{n} (2 \cdot 3 \cdot 4 \cdot . . . \cdot 2006) [∵ \log_{a} (b) + \log_{a} (c) = \log_{a} (b \cdot c)] = \log_{2006!} (2006!) [∵ n = 2006!] = 1 [∵ \log_{a} (a) = 1]$
Hence, option(D) i.e. $1$ is correct

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If n=2006! then 1log2n+1log3n+....+1log2006n=

The correct option is D 1Explanation for correct option Given n=2006!∴1log2n+1log3n+....+1log2006n=logn2+logn3+..+logn2006∵1logab=logba=logn2·3·4·...·2006∵logab+logac=logab·c=log2006!2006!∵n=2006!=1∵logaa=1Hence, option(D) i.e. 1 is correct

If $n = 2006!$ then $\frac{1}{\log_{2} (n)} + \frac{1}{\log_{3} (n)} + . . . . + \frac{1}{\log_{2006} (n)} =$