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

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

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

We move all terms to the left:

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


Domain of the equation: 2)n(n+1)!=0

n∈R



Domain of the equation: 2)(n+1))(n+2))!=0

n∈R


We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+n^2+1/2*1))(n+2))+(+1/2)n(n+1)=0

We calculate fractions


-((+n^2+1*2)nn/2n+2n/2n+1)+(=0


We calculate terms in parentheses: -((+n^2+1*2)nn/2n+2n/2n+1), so:

(+n^2+1*2)nn/2n+2n/2n+1

Fractions to decimals

(+n^2+1*2)nn/2n+1+1

We multiply all the terms by the denominator


(+n^2+1*2)nn+1*2n+1*2n

We multiply parentheses

n^3+2n+1*2n+1*2n

Wy multiply elements

n^3+2n+2n+2n

We do not support enpression: n^3