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
