Question

48+49+....+75=
53+54+55+...+79=
1+3+5+7+....+101=
100 x 99 - 99 x 98+96 x 97 - 97 x 94+95 x 94 - 95
2+4+6...+4008 - (2+6+...+4006)=

Answer 1

48+49+...+75-(1+2+3+...47)+(1+2+3+...+47)=(1+2+3+...+75)-(1+2+3+...47)=(75*76:2)-(47*48:2)=2850-1128=1722

53+54+...79=53+54+...+79-(1+2+3+...52)+(1+2+3+...52)=(1+2+3+...79)-(1+2+3+...+52)=(79*80:2)-(52*53:2)=3160-1378=1782

1+3+5+7+...+101=1+2+3+...+(2*50+1)=51²

100x99-99x98+96x97-97x94+95x94-95=99x(100-98)+97x(96-97)+95x(94-1)=99x2+97x3+95x93=198+291+8835=9324

folosim formulele 1+2+3+...+n=n*(n+1):2
1+3+5+...+(2n+1)=(n+1)²

2+4+6+...+4008-(2+6+...4006)=4007+4008=8015

la ultima  observam ca avem 2-2,4-4....si tot asa pana raman doar  4007 si 4008