If n = 1983! , then value of expression frac{1}{{{{log }₂}n}} + frac{1}{{{{log }₃}n}} + frac{1}{

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Basic of Logarithms

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If $n = 1983!$, then the value of expression $\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + ....... + \frac{1}{{{{\log }_{1983}}n}}$ is equal to

$-1$

$0$

$1$

$2$

(c) $\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + ……. + \frac{1}{{{{\log }_{1983}}n}}$

= ${\log _n}2 + {\log _n}3 + {\log _n}4 + …… + {\log _n}1983$

= ${\log _n}(2.3.4….1983) = {\log _n}(1983!) = {\log _n}n = 1$.

Standard 11

Mathematics

hard

normal

hard

hard

medium