Question

Sa se demonstreze urm.egalitati:     
$\frac{1}{log_2N} + \frac{1}{log_3N}+...+ \frac{1}{log_2_0_0_0N} =1; ~unde~N=1*2*3*...*2000$

Answer 1

$\displaystyle \frac{1}{log_ab}=log_ba \\ \\ log_ax_1+log_ax_2+...+log_ax_n=log_a(x_1x_2...x_n) \\ \\ --------------------- \\ \\ \frac{1}{log_2N}+ \frac{1}{log_3N}+...+ \frac{1}{log_{2000}N}= \\ \\ =log_N2+log_N3+...log_N{2000}= \\ \\ =log_N(2 \cdot 3 \cdot.... \cdot 2000)= \\ \\ = log_NN= \\ \\ =1.$