the sequence $\{a_n\}$ is defined by recursively as $a_0=3$ and $a_n=6a_{n-1}+2$ for all $n\geq0$

Use the iteration to make an educated guess at an explicit formula for a sequence

My attempt: $a_0=3$

then $a_1=6a_0+2=20$



adding all the terms implies

3+20+122+734+.....+(upto n terms) I am stuck form here find an explicit formula for a sequenc

  • $\begingroup$ What have you done so far? Where are you stuck? $\endgroup$ – астон вілла олоф мэллбэрг Nov 7 '16 at 23:52
  • $\begingroup$ @астонвіллаолофмэллбэрг...when i was adding all these i am stuck to find a series of exolict formula $\endgroup$ – user271336 Nov 7 '16 at 23:53
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    $\begingroup$ It looks to me as if you don’t understand what’s meant by the method of iteration. This answer shows an example of it applied to a sequence of the same general kind as yours. $\endgroup$ – Brian M. Scott Nov 7 '16 at 23:55
  • $\begingroup$ @BrianM.Scott...Thanks i got it $\endgroup$ – user271336 Nov 7 '16 at 23:58

Study the sequence $a_n - a_{n-1}$. It goes like $17, 102, 612$ etc. Do you see that this is a geometric series, with first term $17$ and constant ratio $6$?

Hence, $a_n- a_{n-1} = 17 \cdot 6^n$, you could say? Now figure out the general formula for $a_n$, and see if the additions work out. I'l give the answer in this "hidden hint":

$a_n = \dfrac{17\cdot 6^n - 2}{5}$. The sum, up to $n$ terms, is $\dfrac{17\cdot 6^{n+1} -10n-27}{25}$.



Look at the recursion -- it kind of looks like each term is about six times the previous term. So factor out that behavior: Let $a_n = 6^nb_n$. (And since $a_0 = 3$, $b_0 = 3$.)

Your new equation becomes $$ 6^n b_n = 6\cdot6^{n-1} b_{n-1} + 2 \\ b_n = b_{n-1} + \frac{2}{6^{n}} $$

Easier now?

  • $\begingroup$ .....thanks for your help $\endgroup$ – user271336 Nov 8 '16 at 0:28

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