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It says in the solution that the generating function is:

$j=n-2$

$$x^2 \sum_{j=0}^\infty(2j+5)2^jx^j=2x^2\sum_{j=0}^\infty(j+1)(2x)^j+3x^2\sum_{j=0}^\infty(2x)^j=\frac{2x^2}{(1-2x)^2}+\frac{3x^2}{1-2x}$$

I don't understand the these steps. Is there some kind of summation rule that has been applied? Is there any other solution that is easier to follow?

doobiboo
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1 Answers1

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The second sum is just a geometric sequence.

The first sum might me calculated like this

$$2x^2 \sum_{j=0}^{\infty}(j+1)(2x)^j =x^2 \sum_{j=0}^{\infty}\left( (2x)^{j+1} \right)’= x^2 \left(\sum_{j=0}^{\infty}(2x)^{j+1}\right)’ =x^2 \left(\frac{1}{1-2x}\right)’$$

Ilya68
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  • Thanks for your answer! Ok then I understand the second sum. Not sure I understand the above. Is the $x^2$ left out intentionally and what happens with the j variable in $(j+1)$?. Lastly, how do you get the two sums from the leftmost sum? – doobiboo Feb 23 '18 at 18:26
  • @doobiboo I’ve edited the answer. Hope now it’s all clear with $x^2$ and $(j+1)$(it has to be, if you familiar with derivatives). The two sums are obtained by simple representing $2j+5$ as $2(j+1) + 3$ – Ilya68 Feb 24 '18 at 10:30
  • Ok, then i think im with you. Thanks! – doobiboo Feb 26 '18 at 11:33