Question

If n=1983!, compute the sum
$\frac{1}{{\log }_2.n}+\frac{1}{{\log }_{3}n}+\frac{1}{{\log }_{4}n}+\cdots +\frac{1}{{\log }_{1983}n}.$

Answer 1

Since ${\log }_{b}a=\frac{1}{{\log }_{a}b},$ the expression
$={\log }_{n}2+{\log }_{n}3+{\log }_{n}4+......+{\log }_{n}1983$
$={\log }_n\left(\mathrm{2.3.4..........1983}\right)={\log }_{n}1983!$
$={\log }_{n}n=1.[\because 1983!=n]$
Ans: 1