F(n)=F(n-1)+F(n-3)+F(n-5)

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Solution for F(n)=F(n-1)+F(n-3)+F(n-5) equation: 



(F)=(F-1)+(F-3)+(F-5)

We move all terms to the left:

(F)-((F-1)+(F-3)+(F-5))=0



We calculate terms in parentheses: -((F-1)+(F-3)+(F-5)), so:

(F-1)+(F-3)+(F-5)

We get rid of parentheses

F+F+F-1-3-5

We add all the numbers together, and all the variables

3F-9

Back to the equation:

-(3F-9)



We get rid of parentheses

F-3F+9=0

We add all the numbers together, and all the variables

-2F+9=0

We move all terms containing F to the left, all other terms to the right

-2F=-9

F=-9/-2

F=4+1/2

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