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Which one of the following is a closed form expression for the generating function of the sequence $\{a_n\}$, where $a_n = 2n +3 \text{ for all } n=0, 1, 2, \dots$?

  1. $\frac{3}{(1-x)^2}$
  2. $\frac{3x}{(1-x)^2}$
  3. $\frac{2-x}{(1-x)^2}$
  4. $\frac{3-x}{(1-x)^2}$
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@sauravgahlawat can u plz explain how r u getting 5? 

r u putting x=1 in options? cz like tht it’s not possible in a1

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@MANSI_SOMANI Differentiate all options once and then put x = 0 for $a_{1}$. General formula for $a_{n}$ is given above.

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11 Answers

2 votes
2 votes
Let the generating function G(x) for ${a_{k}}$ be $\sum_{k=0}^{\infty }a_{k}.x^k$

$G(x)=\sum_{k=0}^{\infty }a_{k}.x^k$                    $........(1)$

multiply both side  $x$, we get

$x.G(x)=\sum_{k=0}^{\infty }a_{k}.x^{k+1}$         $............(2)$

$1−2$

($1−x).G(x)=\sum_{k=0}^{\infty }a_{k}.x^k - \sum_{k=0}^{\infty }a_{k}.x^{k+1}$

                    $=a_{0}+\sum_{k=1}^{\infty }a_{k}.x^k - \sum_{k=1}^{\infty }a_{k-1}.x^{k}$

                    $=a_{0}+ \sum_{k=1}^{\infty }(a_{k}-a_{k-1}).x^{k}$

                     $=a_{0}+\sum_{k=1}^{\infty }2.x^k $

                     $=a_{0}+2.\sum_{k=1}^{\infty }x^k $

                     $=a_{0}+2x/1-x$

                     $=3+2x/1-x$

                     $= (3-x)/1-x$

$(1-x).G(x)=(3−x)/(1−x)$

          $=(3−x)/(1−x)(1-x)$

          $=(3-x)/(1-x)^2$

 

$N.B$

1.     $a_{k}=2k+3$

         $a_{k−1}=2(k−1)+3=2k−2+3$

         $a_{k}−a_{k−1}=2$

2. sum of infinite $GP$ is $a/1−r $, where $r$ is common ratio and $a$ is the first term

3. $a_{0}=3$
2 votes
2 votes
1 vote
1 vote
Solution by generating functions

edited by
1 vote
1 vote

On evaluating D we get 3*(1-x)-2 - x(1-x)-2  = 3(1+2*x+3*x2+4*x3 ..... so on) - x(1+2*x+3*x2.... so on)

On solving this  3+5x+ 7x2 .... so on, which is the required generating function.

Ans: D

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moved by

Using the basics of Generating function

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nice
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