Question

a) 1+2+3+...........+80






b) 2+4+6+..........+100




c) 1+3+5........+99





d) 3+7+11+15+...........+43



















suma lui gauss

Answer 1

$\displaystyle a).1+2+3+...+80= \frac{80(80+1)}{2} = \frac{80 \times 81}{2} = \frac{6480}{2} =3240 \\ \\ b).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not 2 \times \frac{2550}{\not2} =2550$

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

$\displaystyle d).3+7+11+15+...+43 \\ 3=1 \times 4-1 \\ 7=2 \times 4 -1 \\ 11=3 \times 4-1 \\ 15=4 \times 4-1 \\ . \\ . \\ . \\ 43=4 \times 11-1 \\ 3+7+11+15+...+43=4(1+2+3+4+...+11)-1 \times 11 \\ \\ 3+7+11+15+...+43=4 \times \frac{11 \times 12}{2} -11 \\ \\ 3+7+11+15+...+43=4 \times \frac{132}{2} -11 \\ \\ 3+7+11+15+...+43=4 \times 66-11 \\ \\ 3+7+11+15+...+43=264-11 \\ \\ 3+7+11+15+...+43=253$