Ace Test series: Combinatory - Generating Functions

Question

Answer is B as given in solution.

Comments

Great soln @ankitgupta.1729 Sir.

I have noticed that using differentiation, most of the generating function qs can be easily solved fast. Just like GATE 2022 q. 

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@Abhrajyoti00, Yes. Depends on the question, you could use differentiation, integration, shifting, convolution, Taylor’s approximation etc to solve questions related to generating functions.

For this questions, all the above methods can be used but if you want to solve it fast then you can use Taylor’s approximations, just find the first derivative of all the options and put it zero for $x$, only option (B) will give you $6$ which is nothing but the second number from the given sequence $<0,6,24,...>$  

Another way to solve it fast, is using the fact that $\frac{1}{(1 – x)^c}$ is a closed form for the sequence $<1,c,\binom{c+1}{2},\binom{c+2}{3},... >$. So, here sequence is $<0,6,24,60,..>$ which can also be written as $<0,6*1, 6*4, 6*\binom{4+1}{2},..>$. So, given sequence is nothing but for $c=4,$ it is  shifted one position right and multiplied by $6$, Hence you just have to multiply by $6x$ to $\frac{1}{(1 – x)^4}.$

I would say, if you are not getting any way to solve questions related to generating function and options are available then should go with Taylor’s Approximation. 

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Absolutely Best! Wow! Thanks Sir :))))

Answer 1

ANSWER: OPTION (B) $S_r = \frac{6.x}{(1-x)^4}$

Let, $S_r   =   3.2.1x   +   4.3.2x^2    +    5.4.3x^3    +    6.5.4x^4   + ….  $        

$S_r.x =  \ \ \ \ \ \\ \ \ \ \ \ \\ \ \ \ \ \\ \ \ \  3.2.1x^2   +   4.3.2x^3    +    5.4.3x^4 + …. $      {multiplying both sides with $x$}

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$S_r-S_rx=  6.x+     18.x^2     +    36.x^3   +    60.x^4 +….$ {Subtracting}

$S_r(1-x) =   6.x(1+3.x  +6.x^2  +10.x^3$

$S_r=  (6.x/1-x )  *   (1+3.x  +6.x^2  +10.x^3$

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$S_r = (\frac{6.x}{1-x}).(1+3x + 6.x^2+10.x^3+….)$ → $eqn(1)$

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Let $S1 = 1+3x + 6.x^2+10.x^3+….$

$S1.x =   \ \ \  \ \ \ \ \ \ 1.x+3.x^2 +6.x^3 + \ ...$

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$S1(1-x)= 1+2.x+3.x^2+4.x^3+...$ → $eqn(2)$ {Subtracting}

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Let $S2= 1+2.x+3.x^2+4.x^3+...$

$S2.x= \ \ \ \ \ \ \ \ \ 1.x+2.x^2+3.x^3+...$

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$S2(1-x)= 1+x+x^2+x^3+… = \frac{1}{1-x}$ → $eqn(3)$  {Subtracting}

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Hence, $S2 = \frac{1}{(1-x)^2}$

$\implies S1 = \frac{1}{(1-x)^3}$

Thus, $S_r = \frac{6.x}{1-x}. \frac{1}{(1-x)^3} \implies S_r = \frac{6.x}{(1-x)^4}$ 

 

Answer 2

Solve using options so save time