GATE CSE 2017 Set 2 | Question: 47

Question

If the ordinary generating function of a sequence $\left \{a_n\right \}_{n=0}^\infty$  is $\large \frac{1+z}{(1-z)^3}$, then $a_3-a_0$ is equal to ___________ .

Comments

can you please elaborate this sir ?

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@KUSHAGRA गुप्ता

best solution

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Knowing the “Extended Binomial Theorem” makes Generating Function topic extremely easy. Watch the above playlist to learn everything, with Proof & Variations.

Answer 1

$\frac{1+z}{(1-z)^3} = (1+z)(1-z)^{-3}$

$(1-z)^{-3} = 1 + \binom{3}{1}z + \binom{4}{2}z^2 + \binom{5}{3}z^3 + \dots \infty$

${(1+z)(1-z)^{-3} = (1+z)*(1 + \binom{3}{1}z + \binom{4}{2}z^2 + \binom{5}{3}z^3 + \dots \infty)}$

$a_0$ is the first term in the expansion of above series and $a_3$ is the fourth term (or) coefficient of $z^3$

$a_0$ = coefficient of $z^0 = 1$
$a_3$ = coefficient of $z^3 = \binom{5}{3} + \binom{4}{2} = 10 + 6$

${\Rightarrow a_3 - a_0 = 16 - 1 = 15}$

Comments

plz suggest some good resource to read this topic

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This is a good playlist for Discrete u can follow this topic from here or u can follow keneth rosen https://www.youtube.com/playlist?list=PL0862D1A947252D20

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Generating functions 

Answer 2

Ans : 15

Comments

can you explain the step after 'solving'

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Open the brackets and multiply (1+z) to the terms inside. Simple calculation.

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this is the correct formula.

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nice and easy explanation

Answer 3

yes 15 is correct ans 

Answer 4

Given any sequence say (a₀,a₁,a₂,a₃,....) we represent in generating functions as

                        a₀ + a₁.z + a₂.z² + a₃.z³ + a₄.z⁴ + ....

                                        where z is called as indicator variable.

Now given in question that

                        a₀ + a₁.z + a₂.z² + a₃.z³ + a₄.z⁴ + .... 

                                     = (1 + z) / (1 - z)³

                                                 = (1 + z).( 1 / (1-z)³ )

                                     = (1 + z).(1 / 1-z)³ 

                                     = (1 + z).(1 + z + z² + z³ +z⁴ + ....)³

                                     = 1.(1 + z + z² + z³ +z⁴ + ....)³ + z.(1 + z + z² + z³ +z⁴ + ....)³ ==> eqn 1

From eqn 1, I can say that a₀ = coefficent of z⁰ and a₃ = coefficient of z³

        1.(1 + z + z² + z³ +z⁴ + ....)³

                 = 1.(1 + z + z² + z³ +z⁴ + ....).(1 + z + z² + z³ +z⁴ + ....).(1 + z + z² + z³ +z⁴ + ....)                

               coefficient of z⁰ = 1 (Because z⁰ is possible only if 1 is taken in all 3 bracket terms)

               coefficient of z³ = number of ways of choosing (1,z,z²) + (1,1,z³) + (z,z,z) 

                                     = (3 * 2) + (3C₂ ) + 1

                                     = 6 + 3 + 1

                                     = 10      

z.(1 + z + z² + z³ +z⁴ + ....)³

                                    = z.(1 + z + z² + z³ +z⁴ + ....).(1 + z + z² + z³ +z⁴ + ....).(1 + z + z² + z³ +z⁴ + ....)

                  coefficient of z⁰ = 0 (as we have a z outside and z⁰ is not possible.

                  coefficient of z³ = coefficient of z² from bracket terms

                                            = number of ways choosing (1,1,z²) + (1,z,z)

                                            = 3C₂ + 3C2

                                            = 6

a₀ = Total coefficient of z⁰ =  1 + 0 = 1 

a₃ = Total coefficent of z³ = 10 + 6

                                  = 16

          a₃ - a₀ = 16 - 1 = 15.

Comments

Best answer, thanks