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Suppose $j_0 =a $ and $j_1 = (b\times j_0) +c$ and $j_2=(b\times j_1)+c $ and $\cdots j_n =(b\times j_{n-1})+c $. I want a formula for $j $ as a function of $n: j= f (n) $. What is $f $?

If this is possible can you show me the method please?

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    Why not encode the image? – 2017-02-06
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    @ΘΣΦGenSan thank's for the edit, i find it very difficult and confusing to format text on this website – 2017-02-06
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    Well, you need some extra research. You read this http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-06
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    If you want people to help you you will need to show that you can put more effort into putting the question like this. Can you type the question, instead of including a link to a figure? Can you tag your question properly? Please see http://math.stackexchange.com/help/how-to-ask for more. And http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for how to type math. – 2017-02-06
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    Actually i typed it in Wordpad and i took a screen shot, and i uploaded the picture, because i don't know how to write subscripts. – 2017-02-06
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    Try \$j_{0}\$ and \$j_{n-1}\$, etc. – 2017-02-06
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    @FredKline thank's, that's what i call a help (y) – 2017-02-06

3 Answers 3

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We have the formula $$j_n =(bj_{n-1})+c $$ $$=b (bj_{n-2}+c) +c = b^2j_{n-2} + bc +c $$ $$=b^2 (bj_{n-3}+c) + bc +c= b^3j_{n-3} + b^2c + bc +c$$ $$=\cdots $$

We can see that a pattern is emerging as we have $$j_n = b^kj_{n-k} + b^{k-1}c + b^{k-2}c + \cdots + bc +c $$

Now put $k = n $ and see what you get by also substituting $j_0 =a $. Hope it helps.

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    $$j_n = b^nj_{n} + b^{n-1}c + b^{n-2}c + \cdots + bc +c $$ $$j_n = b^nj_{n} + c (b^{n-1} + b^{n-2} + \cdots + b +1) $$ and we have $\sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a}$ (geometric series with starting index i=0) so $$j_n = b^nj_{n} + c \frac{1-b^n}{1-b} $$ @rohan is this correct? – 2017-02-06
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    @MaykelJakson The coefficient of $b^n $ is $j_0=a $, not $j_n $. Otherwise all is fine. – 2017-02-06
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    yes i wanted to write j0 and i wrote jn, thank's for all – 2017-02-06
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    i forgot to mention: for b ≠ 1 – 2017-02-11
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Hint: from your $j_{0}=a$ and $j_{1}=bj_{0}+c$, you can substitute to get $j_{1}=ab+c$. You also have $j_{2}=bj_{1}+c$. Substitutting again you get $j_{2}=ab^2+bc+c$. Can you keep going to $j_{n}$?

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Try to examine inverse iterations. We have $j_n$ as the $n$-th iterate of a linear function.