ACE Test Series: Generating Function

Question

The generating function of the sequence

$\left \{ a_{0},a_{1},a_{2}..........a_{n}………...\infty \right \}$

where $a_{n}=\left ( n+2 \right )\left ( n+1 \right ).3^{n}$ is

$a)3\left ( 1+3x \right )^{-2}$

$b)3\left ( 1-3x \right )^{-2}$

$c)2\left ( 1+3x \right )^{-3}$

$d)2\left ( 1-3x \right )^{-3}$

Answer 1

Definition of Generating Function says :-

                                                                  $G(x) = \sum_{n=0}^{\infty }a_{n}x^{n}$

Where $a_{n}$ is called the sequence. So, $<a_{0},a_{1},a_{2},......>$ is the sequence for the generating function.

Here, Sequence $a_{n}$ is defined as $(n+2)(n+1)3^{n}$ where $n=0,1,2,3,........$

So,

$a_{0}=2*1*3^{0}$

$a_{1} = 3*2*3^{1}$

$a_{2}= 4*3*3^{2}$

$a_{3}= 5*4*3^{3}$

$a_{4}=6*5*3^{4}$

$......$

$a_{n}=(n+2)*(n+1)*3^{n}$

Now, According to definition of $G(x),$

$G(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} +........+a_{n}x^{n}+............$

$G(x) = 2*1*3^{0} + 3*2*3^{1}x + 4*3*3^{2}x^{2} + 5*4*3^{3}x^{3}+......+(n+2)(n+1)3^{n}x^{n} +......$

$G(x) = 2*1*(3x)^{0} + 3*2*(3x)^{1} + 4*3*(3x)^{2} + 5*4*(3x)^{3}+..$ ---- $Equation (1)$

Now, Multiply the whole equation by $3x$ and write the whole equation be leaving one term. So, It becomes

$3xG(x) =\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2*1*(3x)^{1} + 3*2*(3x)^{2} + 4*3*(3x)^{3} + 5*4*(3x)^{4} +....$

Say, It is $Equation (2)$

Now, subtract equation $(2)$ from equation $(1)$ vertically terms by terms. So, after that we will get :-

$(1-3x)G(x) = 2(3x)^{0} + 4*(3x)^{1} + 6*(3x)^{2} + 8*(3x)^{3}+....$--- $Equation(3)$

Now, again multiply by $3x$ in the whole equation$(3)$ and write it by leaving one term of equation(3), So, It becomes :-

$(1-3x)3xG(x) =\;\;\;\;\;\;\;\;\;\; 2*(3x)^{1} + 4*(3x)^{2} + 6*(3x)^{3} + 8*(3x)^{4}+...$--$Equation(4)$

Now, again subtract equation $(4)$ from equation $(3)$ vertically terms by terms. After that we will get:-

$(1-3x)^{2}G(x) =2(3x)^{0} + 2*(3x)^{1} + 2*(3x)^{2} + 2*(3x)^{3}+.......$

So,

$(1-3x)^{2}G(x) = 2\times \frac{1}{(1-3x)}$

$\Rightarrow G(x) = 2\times \frac{1}{(1-3x)^{3}}$

So, $G(x) = 2(1-3x)^{-3}$

---

Edit (16/03/2022):

We can also use one time saving approach from this comment to eliminate options here.

$a_0 = f(0)$ and $a_1 = f’(0)$

Putting $x=0$ in each option, we get $f(0)$ as

For options a) and  b) $f(0) = 3 $ and For options c) and  d) $f(0) = 2 $

Since $a_0 = 2$, So, Either  (c) is correct or (d) because $a_0 = f(0).$

Now, For check $f’(0)$ for each option

for (c), $f’(0) = -6*3 = -18$

for (d), $f’(0) = -6*-3 = 18$

Since, $a_1= 18 $ and $f’(0) = a_1$.

It means option (d) is correct.

Answer 2

Let the general form generating function be denoted as $b(x) = \sum_{n=0}^{\infty} b_nx^n$. 

One simple observation can get you $3^n$, and that is making $x \rightarrow 3x$. 

$b(3x) = \sum_{n=0}^{\infty}b_n3^nx^n$

Let's do the differentiation wrt $x$ on both sides (why?)

$b'(x) = \sum_{n=1}^{\infty}nb_n3^nx^{n-1} = \sum_{n=0}^{\infty}(n+1)b_{n+1}3^{n+1}x^{n}$

So, $\dfrac{b'(3x)}{3} = \sum_{n=0}^{\infty}b_{n+1}(n+1)3^nx^{n}$  

Let's do the differentiation wrt $x$ on both sides (in hope to get $(n+2)$ this time)

$\dfrac{b''(3x)}{3} = \sum_{n=1}^{\infty}b_{n+1}(n)(n+1)3^nx^{n-1} = \sum_{n=0}^{\infty}b_{n+2}(n+1)(n+2)3^{n+1}x^{n}$  

So, $\dfrac{b''(3x)}{9} = \sum_{n=0}^{\infty}b_{n+2}\underbrace{(n+1)(n+2)3^n}_{a_n}x^{n}$ 

Hence, $(b_n = 1)_{n=0}^{\infty}$ (must be)

We know, for $\boxed{b(x) = (1-x)^{-1}}$, $(b_n = 1)_{n=0}^{\infty}$

So, $\dfrac{1}{9}b''(3x) = \dfrac{1}{9} \dfrac{d^2}{dx^2} \dfrac{1}{(1-3x)} = \dfrac{1}{9} \dfrac{d}{dx} \dfrac{3}{(1-3x)^{2}} = \dfrac{1}{9} \dfrac{18}{(1-3x)^{3}} = 2(1-3x)^{-3}$ 

$\textbf{Option (D) is correct}$

Answer 3

Here, Sequence an is defined as (n+2)(n+1)3n where n=0,1,2,3,........

So,

a0=2∗3^0

a1=6∗3^1

a2=12∗3^2

a3=20∗3^3

a4=30∗3^4

and we know that

1/(1-x) =1+x+x^2+x^3+x^4......(   GP.)

1/(1-3x)=1+(3x)+(3x)^2+(3x)^3...

 

 

Answer 4

Using properties of generating functions, we can greatly reduce the pain stalking naive approach