If n = 1983!, then the value of expression fr | vedclass

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Question

(c) $\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + ....... + \frac{1}{{{{\log }_{1983}}n}}$

= ${\log _n}2 + {\lo

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(c) $\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + ....... + \frac{1}{{{{\log }_{1983}}n}}$

= ${\log _n}2 + {\log _n}3 + {\log _n}4 + ...... + {\log _n}1983$

= ${\log _n}(2.3.4....1983) = {\log _n}(1983!) = {\log _n}n = 1$.

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

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