Question

1+2+3+...+n=210
Aflati n

Answer 1

gauss
1+2+3+...+n=n(n+1)/2

n(n+1)/2=210
n(n+1)=420
n^2+n=420
putem afla n cu delta
n^2+n-420=0
delta=1-4*(-420)
delta=1+1680
delta=1681
radical delta=41

n nu are voie sa fie negativ, deci singura varianta va fi
n=1+41/2
n=42/2
n=21

Answer 2

Aplicam formula lui Gauss:
1+2+...+n=n(n+1)/2
1+2+3+...+n=n(n+1)/2=210
n(n+1)=2x210
n(n+1)=420
n²+n-420=0
Δ=b²-4ac
Δ=1²-4x1x(-420)=1-4(-420)=1+1680=1681
n   nu   ia valori negative,deci:
n=1+√1681/2=1+41/2=42/2=21
deci:n=21