F(n)=F(n-1)+n*2^n

Solution for F(n)=F(n-1)+n*2^n equation:

(F)=(F-1)+F*2^F

We move all terms to the left:

(F)-((F-1)+F*2^F)=0


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

(F-1)+F*2^F

Wy multiply elements

2F^2+(F-1)

We get rid of parentheses

2F^2+F-1

Back to the equation:

-(2F^2+F-1)


We get rid of parentheses

-2F^2+F-F+1=0

We add all the numbers together, and all the variables

-2F^2+1=0

a = -2; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-2)·1
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-2}=\frac{0-2\sqrt{2}}{-4}
=-\frac{2\sqrt{2}}{-4}
=-\frac{\sqrt{2}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-2}=\frac{0+2\sqrt{2}}{-4}
=\frac{2\sqrt{2}}{-4}
=\frac{\sqrt{2}}{-2}
$