Question

Consideram suma S =1+2+3+.......+n n,n apartine lui N .Determinati n daca n+1 reprezinta 4%din suma

Answer 1

$\frac{4}{100}* \frac{n(n+1)}{2}=n+1$

$\frac{4}{100}* \frac{n^2+n}{2}=n+1$

$\frac{4n^2+4n}{200}=n+1$

simplificam membrul stang cu 4 si pe cel drept il amplificam cu 50 ptu a aduce la acelasi numitor

$\frac{n^2+n}{50}= \frac{50n+50}{50}$

$n^2+n=50n+50$

$n^2+n-50n-50=0$

$(n^2+n)-(50n+50)=0$

$n(n+1)-50(n+1)=0$

$(n-50)(n+1)=0$

n-50=0, n=50
n+1=0, n=-1

Deoarece n∈N doar n=50 e solutie a problemei.