If n-1983 - then the value of expression 1-log 2 n-1-log 3 n-1-log 4 n-1-log 1983 n is equal toA- -1B- 0C- 1D- 2

The correct option is C 1log_n 2+log_n3+… log_n 1983 =log_n[2.3.4…1983]=log_n 1983! =log_n n=1 ∵ 1983!=n ...

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Fundamental Laws of Logarithms

If n=1983 !, ...

Question

If n=1983!, then the value of expression $\frac{1}{l o g_{2} n} + \frac{1}{l o g_{3} n} + \frac{1}{l o g_{4} n} + \dots \frac{1}{l o g_{1983} n}$ is equal to

-1

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\frac{1}{l o g_{2} n} + \frac{1}{l o g_{3} n} + \frac{1}{l o g_{4} n} + \dots \frac{1}{l o g_{1983} n} .

\frac{1}{{log}_{2} . n} + \frac{1}{{log}_{3} n} + \frac{1}{{log}_{4} n} + \dots + \frac{1}{{log}_{1983} n} .

n > 1

\frac{1}{{log}_{2} n} + \frac{1}{{log}_{3} n} + \frac{1}{{log}_{4} n} + . . . . . . + \frac{1}{{log}_{50} n}

\frac{1}{l o g_{2} n} + \frac{1}{l o g_{3} n} + \frac{1}{l o g_{4} n} + \dots \frac{1}{l o g_{1983} n}

\frac{1}{{log}_{2} n} + \frac{1}{{log}_{3} n} + \frac{1}{{log}_{4} n} + . . . . . . . . . . . . . + \frac{1}{{log}_{43} n}

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