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

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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

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