Question

calculati urmatoarele numere 1+2+3+...+80,2+4+6+...+100,1+3+5+...+99,3+7+11+15+...+43

Answer 1

Singura formula de care ai nevoie pentru a calcula aceste sume este cea a lui Gauss: 1+2+3+...+n=n*(n+1)/2, oricare ar fi n numar natural.
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1+2+3+...+80=80*81/2=40*81=3240.

2+4+6+...+100=2*(1+2+3+...+50)=2*50*51/2=50*51=2550.

1+3+5+...+99=(1+2*0)+(1+2*1)+(1+2*2)+...+(1+2*49)=50+2*(1+2+3+...+49)=50+2*49*50/2=50+49*50=50+2450=2500.

3+7+11+15+...+43=(3+4*0)+(3+4*1)+(3+4*2)+(3+4*3)+...+(3+4*10)=11*3+4*(1+2+3+...+10)=33+4*10*11/2=33+2*10*11=220+33=253.