Question

1+3+5+.....+99 Rezolvarea va rog fiindca nu inteleg

Answer 1

1+3+5+...+99
Observăm că avem suma de numere impare consecutive. Vom scrie fiecare număr de forma asta:
1=2•0+1
3=2•1+1
5=2•2+1
7=2•3+1
...
99=2•49+1
1+3+5+7+...+99=
=(2•0+1)+(2•1+1)+(2•2+1)+(2•3+1)+...+(2•49+1)=
=2•0+2•1+2•2+2•3+...+2•49+1+1+1+1+...+1(de 50 de ori)=
(îl dăm pe 2 factor comun)
=2(1+2+3+...+49)+50=
=2•49•50/2+50=
=49•50+50=
(îl dăm pe 50 factor comun)
=50(49+1)=
=50•50=2500
;))

Answer 2

1=2×1-1 3=2×2-1 5=2×3-1 . . . 99=2×50-1 1+3+5+.....+99=2×(1+2+3+....+50)-1×50 1+2+3+....+50=(50+1)×50÷2= =51×50÷2= =51×25= =1275 1+3+5+......+99=2×1275-50= =2550-50= =2500