Question

ma puteti ajuta cu aceste exerciti,va rog; Calculati: a) 2·(1+2+3+.....+100)·(1/1·2+1/2·3+1/3·4+.....+1/100·101);                                                      b) [2002-(1/2+2/3+3/4+....+2002/2003)]÷(1/2+1/3+1/4+....+1/2003)

Answer 1

a) A= 2·100·101/2 =100·101       B= 1/1-1/2 +1/2-1/3+1/3-1/4+ + ...... +1/100 - 1/101 =1/1 -1/101=100/101

A·B= 100·101·100/101 =100² =10000

b) S=1/2+2/3+3/4+....+2002/2003 = (1- 1/2)+(1-1/3)+(1-1/4)+........+ (1-1/2003) = 1·2002 - (1/2+1/3+1/4+........+1/2003)

 2002 - S = - (1/2+1/3+1/4+......+1/2003)

S1 =1/2+1/3+1/4+......+1/2003 ⇒(2002 - S)÷S1 = - 1