Question

Calculați (1*2+2*3+3*4+....+2002*2003+2003*1002):(1^2+2^2+3^2+....2003^2)

Answer 1

Folosim sume cunoscute.

Answer 2

Ne vom ocupa de suma din prima paranteza, exceptand ultimul termen.

1·2+2·3+3·4+ ... +2002 ·2003=

=1(1+1) + 2(2+1) + 3(3+1)+ ... +2002(2002+1) =

=1+1+2²+2+3²+3 + ... +2002²+ 2002 =

=(1²+2²+3²+ ... +2002²) + (1+2+3+ ... +2002) =

=(1²+2²+3²+ ... +2002²) +(2002·2003)/2 =

=(1²+2²+3²+ ... +2002²) + 1001·2003

Adaugam acum ultimul termen al sumei si obtinem:

(1²+2²+3²+ ... +2002²) + 1001·2003 + 2003·1002 =

=(1²+2²+3²+ ... +2002²) + 2003(1001+1002) =

=(1²+2²+3²+ ... +2002²) + 2003² = 1²+2²+3²+ ... +2002² + 2003²

Asadar, exercitiul devine:

(1²+2²+3²+ ... +2002² + 2003²):(1²+2²+3²+ ... +2002² + 2003²) =1