Question

Calculati media aritmetica a numerelor a=1pe2+1pe3+...+1pe1999 iar b=1pe2+2pe3+...+1998pe1999. PLS!!!
Ofer 54 de PUNCTE!!!

Answer 1

[tex]a=\frac{1}{2}+\frac{1}{3}+_{\dots} +\frac{1}{1999}\\ \\ b=\frac{1}{2}+\frac{2}{3}+_{\dots}+\frac{1998}{1999}\\ m_a=\frac{a+b}{2}\\ m_a=\frac{\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)+_{\dots}+\left(\frac{1}{1999}+\frac{1998}{1999}\right)}{2}\\ ~~~~~~~~~~~~de\ 1998\ ori\\ m_a=\frac{\overbrace{1+1+_{\dots}+1}}{2}\\ m_a=\frac{1998}{2}=\boxed{999}[/tex]