Question

sume gauss 1+3+5+........+99

Answer 1

$\displaystyle 1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2 (1+2+3...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950- \not 2 \times \frac{2450}{\not 2} =4950-2450=2500$