Question

37+41+45+...+121=?
:-)

Answer 1

Scadem din fiecare 37 .... si rezulta 0+4+8+....+ 84.......Acum dam factor comun pe 4 si rezulta  4( 0+1+2+...+21)....de mai sus deducem ca am avut 22 numere din care l-am scazut pe 37 ..... 4(1+2+...+21)+ 22*37 = 4 ($\frac{(1+21)*21}{2}$) + 814 = 924+814 =1738

Answer 2

1.37+41+45+...+121=
(37+41+45+...+121)- (33+29+25+...+ 1)=
[(36+ 1)+(40+ 1)+(44+ 1)+ ...+(120+ 1)]- [(32+ 1)+(28+ 1)+(24+ 1)+ ...+(0+ 1)]=
[(1+ 1+ 1+ ...+1)+(36+40+44+ ...+120)]- [(1+1+1+ ...+1)+(32+28+26+ ...+ 0)]=
[1·22+(4·9+4·10+4·11+...+4·30)]-[1·9+(4·8+4·7+4·6+ ...+4·0)]=
[22+4·(9+10+11+ ...+30)]-[9+4·(8+7+6+ ...+0)]=
(22+4·30·31:2)-(9+4·8·9:2)=
[22+(4·15·31)]- (9+4·4·9)=
(22+60·31)-(9+16·9)=
(22+1 860)-(9+144)=
1 882-  153=
1 738