Question

Suppose $f\left(n\right)={\log }_2\left(3\right).{\log }_3\left(4\right).{\log }_4\left(5\right).......{\log }_{n-1}\left(n\right)$ then the sum ${\sum }_{k=2}^{100}f\left(2^k\right)=\lambda$,

then $\frac{\lambda }{1683}$ is equal to?

Answer 1

$f\left(n\right)=lo{g}_2\left(3\right).lo{g}_3\left(4\right).lo{g}_4\left(5\right).......lo{g}_{n-1}=$$lo{g}_{2}n$

now ${\sum }_{k=2}^{100}f\left(2^k\right)=lo{g}_2\left(2^2\right)+lo{g}_2\left(2^3\right)+lo{g}_2\left(2^4\right)+.....+lo{g}_2\left(2^{100}\right)$

$=lo{g}_2(2^2{.2}^3-2^4{.2}^5....{.2}^{100})$

$=lo{g}_2\left(2^{2+3+4+\mathrm{5.....}+100}\right)$

$=(1+2+3+4+.....+100)-1$

$=\frac{100\left(101\right)}{2}-1$

$\lambda =5050-1=5049$

$\therefore \frac{\lambda }{1683}=\frac{5049}{1683}=3$