Question

Se considera numerele rationale

x = $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{2002} + \frac{1}{2003}$

y = [tex] 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +...+ \frac{1}{2001} - \frac{1}{2002}

a) Calculati media aritmetica a numerelor x si y

b) Aratati ca x < \frac{1001}{2003} < y

Va rog ajutati ma urgent, maine am teza la matematica!!!

Answer 1

a. x+y /2=
1-1/2+1/2-1/3+1/3-1/4+1/4-...+1/2003 /2 =
1-1/2003 /2 =
2002/2003/2=
2002/4006=1001/2003 (media aritmetica a celor 2 numere);
b. Media aritmetica a celor doua numere x si y este 1001/2003 . Atunci suma celor doua numere va fi x+y=2002/2003.
x+y=2002/2003
Stiind ca x<y ⇒x<1001/2003<y .