Question

Calculati urmatoarele sume 1+3+5+...+99 va rog frumos sa ma ajutati .

Answer 1

$\displaystyle 1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950- \not 2 \times \frac{2450}{\not2} =4950-2450=2500$

Answer 2

S=1+2+3+...+99
S=99+97+...+1
2S=100+100+...+100
S=100•50/2
S=50•50
S=2500
(sunt 50 de numere impare pana la 99)