-3(1/3)n+1/3+1(1/2)n=-5(1/6)

Solution for -3(1/3)n+1/3+1(1/2)n=-5(1/6) equation:

-3(1/3)n+1/3+1(1/2)n=-5(1/6)

We move all terms to the left:

-3(1/3)n+1/3+1(1/2)n-(-5(1/6))=0


Domain of the equation: 3)n!=0

n!=0/1

n!=0

n∈R



Domain of the equation: 2)n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

-3(+1/3)n+1(+1/2)n+1/3-(-5(+1/6))=0

We multiply parentheses

-3n^2+n^2+1/3-(-5(+1/6))=0

We calculate fractions


-3n^2+n^2+()/()+()/()=0

We add all the numbers together, and all the variables

-2n^2+2=0

a = -2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-2)·2
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-2}=\frac{-4}{-4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-2}=\frac{4}{-4}
=-1
$