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

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

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

$a_0$ 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 $z^3 = \binom{5}{3} + \binom{4}{2} = 10 + 6$

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

10 votes
10 votes

yes 15 is correct ans 

10 votes
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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)

                                     = (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.

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