Question

1+3+5+7+9+...+99 = ?

Answer 1

1+3+5+7+9+...+99=

= 1+2+3+...+99 - ( 2+4+6+...+98)=

$= \frac{99 \times 100}{2} - 2 \times (1 +2 + ... + 49) =$

$= 99 \times 50 - 2 \times \frac{49 \times 50}{2} =$

$= 99 \times 50 - 49 \times 50 =$

$50\times (99 - 49) =$

$= 50 \times 50 = {50}^{2}$

Answer 2

Salut,

1+3+5+7+9+......+99=

=1+2+3+....+99-(2+4+6+...+98)=

=99×100/2-2×(1+2+....+49)=

=99×50-2×49×50/2=

=99×50-49×50=

=50×(99-49)=

=50×50=

=50^2=100

$by \\ 7646$