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Question Ok so i have to solve the question above and find the general term of this sequence without using the induction method or the countdown method. So i want to calcul this with the geometric formula or arithmethic given : formula

I thought this would be a summation of an arithmetic sequence but using the formula i don't get the answer.

I've tested the answer with wolframalpha entering : y(0)=2,y(n)=2y(n-1)+3n-4

Using the geometric formula i get close to the answer but missing -3n, so i'm kinda lost. Could anyone help thanks !

Maybe it can't be solved with geometric or arithmetic formula and i'd have to use an other method ?

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2 Answers 2

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Assume $a_n=b_n+pn+q$ therefore $$$$

therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\\=2b_{n-1}+3n-4+pn+q\\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$

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  • $\begingroup$ First, thank you for your answer, but i'm kinda unsure what made you define $a_n$ like that ? Also why did you replace it there in the $b_n$ thank you ! $\endgroup$
    – Dany Pépin
    Commented Nov 29, 2018 at 12:53
  • $\begingroup$ You're welcome. This is because a linear term like $3n-4$ strongly should have a linear source too (in some cases this is not possible). This made me think that $a_n=b_n+pn+q$ can be such that the representation of $a_{n+1}$ as a linear function of its priors, does not depend on $n$ (or any linear combination of n such as $3n-4$ in this case). By a simple calculation, I derived $p$ and $q$ and substituted. Also once you could simplify $b_n$ to $a_n$ you'd be better substitute it anywhere with $a_n$ to make your proof easier. $\endgroup$
    – Mostafa Ayaz
    Commented Nov 29, 2018 at 13:00
  • $\begingroup$ Wait, you used derivative to obtain $a_n$ ? not sure we saw that in class haha, how would you obtain it without it ? I'm unsure. thanks ! $\endgroup$
    – Dany Pépin
    Commented Nov 29, 2018 at 13:44
  • $\begingroup$ I didn't say that! Derivation is not allowed in concrete mathematics..... $\endgroup$
    – Mostafa Ayaz
    Commented Nov 29, 2018 at 13:46
  • $\begingroup$ Then i'm not quite sure how you landed on $b_n$ $+3n +2$ hahaha $\endgroup$
    – Dany Pépin
    Commented Nov 29, 2018 at 14:14
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Let $b_m=a_m+p+qm$

$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$

Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$

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  • $\begingroup$ What i've done : $2(1-2^{n+1})/(1-2)$ which result in $2^{n+2}-2$ seems like i have the same answer as you but from wolframalpha, the answer should be : $b_n =-3n +2^{n+2} - 2$ $\endgroup$
    – Dany Pépin
    Commented Nov 28, 2018 at 16:06

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