1+2+3+4+n=n(n+1)/n

Solution for 1+2+3+4+n=n(n+1)/n equation:

1+2+3+4+n=n(n+1)/n

We move all terms to the left:

1+2+3+4+n-(n(n+1)/n)=0


Domain of the equation: n)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

n-(n(n+1)/n)+10=0

We multiply all the terms by the denominator


n*n)-(n(n+1)+10*n)=0

Wy multiply elements

n^2+10n=0

a = 1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*1}=\frac{-20}{2}
=-10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*1}=\frac{0}{2}
=0
$