solve the following using generating function

Question

$a_n-a_{n-1} =3(n-1), n\ge1$ and where $a_0=2$ how should I solve this? I am getting, $F(x)=3(2x^2-3x+1)/(1-x)^2(1-x)$ while doing partial fraction I am getting $A+B+C=1, 2A+2B+2C=2 A+C=3$. Unable to solve from this portion.

Answer 1

Actually, the generating function is$$F(x)=\frac{3x^2}{(1-x)^3}.$$Now, since$$\frac1{1-x}=1+x+x^2+x^3+\cdots,$$we have$$\frac1{(1-x)^2}=\left(\frac1{1-x}\right)'=1+2x+3x^2+4x^2+\cdots$$and so$$\frac1{(1-x)^3}=\frac12\left(\frac1{(1-x)^2}\right)'=1+3x+6x^2+10x^3+\cdots+\frac{(n+1)(n+2)}2x^n+\cdots$$It follows that$$F(x)=\sum_{n=1}^\infty\frac32(n-1)nx^n.$$

Answer 2

Starting from the point you got to, using partial fractions method, you should get: $$\frac{3(2x^2-3x+1)}{(1-x)^3}=\frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{(1-x)^3}$$ Giving you: $$3(2x^2-3x+1)=A(1-x)^2+B(1-x)+C=-Ax^2-(2A+B)x+A+B+C$$