Question

2070-2069+2068-2067+...+4-3+2-1
1+2+3+4+...+n
N = ? Ma puteti ajuta

Answer 1

Formula 1+2+3+...+n=[n(n+1)]/2

S=(2070-2069)+(2068-2067)+(2066-2065)+...+(4-3)+(2-1)

S=1+1+1+...+1+1

2070/2=1035 termeni, grupați câte 2

S=1*1035

S=1035

S=n(n+1)/2

n(n+1)=2S

n(n+1)=2*1035

n(n+1)=5*3^2*2*23

n(n+1)=45*46

n=45

n+1=45+1

n+1=46

Answer 2

$\frac{2070 - 2069 + 2068 - 2067 + .... + 4 - 3 + 2 - 1}{1 + 2 + 3 + 4 + .... + n}$

$(2070 - 2069) + (2068 - 2067) + ..... + (2 - 1) = 1 \times \frac{2070}{2} = 1035$

$numitorul \: este \: \: \frac{n \times (n + 1)}{2}$

$daca \: \frac{n \times (n + 1)}{2} = 1035 \: atunci \: n \times (n + 1) = 2070 = 2 \times {3}^{2} \times 5 \times 23 = 45 \times 46 \: \: \: deci \: = > n = 45$