Find generating function created by $a_n=2^{n+3}$, for $n\geq3$,$ a_0=1$, $a_1=\frac{5}{2}$ $a_2=3$

Question

I hope title is understandable, wasn't sure on how to translate this task from my language.

Below is my solution to this problem, is logic behind it correct?

$$A(z)=\sum_{n=0}^{\infty}a_nz^n=1+\frac{5}{2}z+3z^2+\sum_{n\geq3}^{\infty}a_nz^n$$ $$=1+\frac{5}{2}z+3z^2+2^3+\sum_{n\geq3}^{\infty}(2z)^n$$ $$=1+\frac{5}{2}z+3z^2+2^3+[(2z)^3+(2z)^4+(2z)^5...]$$

$$b_n=(2_z)^n$$

$$\frac{b_{n+1}}{b_n}=2z$$

$$q=2z$$ $$W_0=(2z)^3$$ $$|2z|<1|$$ $$=1+\frac{5}{2}z+3z^2+2^3+\frac{(2z)^3}{1-2z}$$

I need to make sure on test that I didn't make mistakes so if its correct, could any of you explain on how to check if answer is right?

You haven't used $a_0=19$, and you have magicked a value for $a_2$ out of nowhere. — Gerry Myerson, Feb 05 '19 at 12:23
You have $2^3$ plus summation, that should be $2^3$ times summation. — Gerry Myerson, Feb 06 '19 at 00:50

score 2 · Answer 1 · answered Feb 05 '19 at 12:03

2

I think your work is wrong: $$\begin{align} A_n(x)&=\sum_{n \in \mathbb {N}}a_n x^n\\ &=a_0x^0+a_1x^1+a_2x^2+\sum_{n \geqslant3}a_n x^n\\ &=19+x+a_2x^2+\sum_{n \geqslant 3}2^{n+3}x^n\\ &=19+x+a_2x^2+2^3\sum_{n \geqslant 3}(2x)^n\\ \end{align}$$

answered Feb 05 '19 at 12:03

Botond

11,938

Could you check again? I made an error while typing title, perhaps now my solution is correct. – Gorosso Feb 05 '19 at 22:03
1

@Gorosso Sorry, somehow I missed the notification. $2^3$ will multiply the sum, as you can see in my work. But the rest seems to be fine. – Botond Feb 07 '19 at 00:03

Find generating function created by $a_n=2^{n+3}$, for $n\geq3$,$ a_0=1$, $a_1=\frac{5}{2}$ $a_2=3$

1 Answers1