1

$a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$

So I calculate first terms

$a_0=0$

$a_1=-6$

$a_2=-24$

$a_3=-78$

I don't see any relation so

$a_n=a_{n-1}-6\cdot3^{n-1}$

$a_{n-1}=a_{n-2}-6\cdot 3^{n-2}$

. . .

$a_2=a_1-6\cdot3^{1}$

$a_1=a_0-6\cdot 3^{0}$

Not sure what to do next, Wolfram solves it in this way:

$a_n=-3\cdot(3^{n}-1)$

How do I get to this point?

Stefan4024
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Gorosso
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4 Answers4

2

You have $$a_n=a_n=a_{n-1}-6\cdot 3^{n-1}=a_{n-2}-6\cdot 3^{n-2}-6\cdot 3^{n-1}=\dotsm,$$ so you can prove with an easy induction that $$a_n = a_0-6\sum_{k=0}^{n-1} 3^k=-6\frac{3^n-1}{3-1}.$$

Bernard
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2

hint...consider $$\sum_{r=0}^{r=n}(a_{r+1}-a_r)=\sum_{r=0}^{r=n}-6\cdot3^r$$

The LHS is a telescoping series and the RHS is a geometric series.

David Quinn
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1

Note that for all $n$ we have $$a_{n+1}-a_n = -6\cdot 3^n $$

so we have also: $$a_n-a_{n-1}=-6\cdot3^{n-1}$$

thus $$a_{n+1}-a_n = 3(-6\cdot3^{n-1}) = 3(a_n-a_{n-1})$$

{or divide this two equations: $${a_{n+1}-a_n \over a_n-a_{n-1}}= {-6\cdot 3^n\over -6\cdot3^{n-1}} = 3$$}

so you have to solve linear recurrence:

$$ a_{n+1}-4a_n+3a_{n-1}=0$$

CAn you do that?

nonuser
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  • ok so:

    $x^{2}-4x+3$

    $\Delta =4$

    $x1=1$ $x2=3$

    Now I use following formula

    $a_n=ar^{n}+br^{n}$

    $a_0=0=a+b$

    $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=31^{n}-33^{n}$

     – Gorosso Jan 27 '19 at 14:40
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    Yes, that is correct! – nonuser Jan 27 '19 at 14:40
  • Yeah, I am getting correct numbers for a1, a2, a3 etc. Though I am still not quite sure how you achieved linear recurrence form, could you expand on that? (Sorry for lack of new lines in above comment, in my original post newline worked correctly) – Gorosso Jan 27 '19 at 14:45
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    Is that better? – nonuser Jan 27 '19 at 14:52
  • I understand whats happening in 2 first lines, unfortunately after that I am lost. $a_(n+1)-a_n=-6*3^{n}$ so how come in the third line first '3' before the bracket appears out of nowhere? – Gorosso Jan 27 '19 at 15:01
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    What about now? – nonuser Jan 27 '19 at 15:51
  • I mean, I understand lines you add, but still got no idea how to get linear recurrence from that later. Is it dumb of me that I can solve linear recurrence but can't mold it in the first place? – Gorosso Jan 27 '19 at 16:07
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    Just put first and last expression on one side, so you get 0 on other side – nonuser Jan 27 '19 at 16:54
  • Ohh I see now that, was silly of me. I might pass my exam tomorrow thanks to you. – Gorosso Jan 27 '19 at 16:57
1

This is just a Geometric Series:

$$a_n=-6\sum_{i=1}^n3^{i-1}=-6\sum_{i=0}^{n-1}3^i=-6\times \frac {3^n-1}{3-1}=-3\times (3^n-1)$$

lulu
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