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A sequence $u_1, u_2, u_3$,... is such that $u_1=1$ and $u_{n+1}=4u_n +7$ for $n \geqslant 1$.

Write down the first four terms of the sequence.

I have solved the first half of the question.

$T_1 =1$

$T_2 =11$

$T_3 =51$

$T_4 =211$

What kind of sequence is this? It can’t be a geometric progression since there is no common ratio, and can’t be an arithmetic progression either since there is no common difference.

I need help on solving the second half of the question.

Show that an explicit formula for $u_r$ is given by $u_r = 1+ \frac {10}{3} [4^{r-1} -1]$

How to show this? Do I use the given formulas in the question? Or is it $u_r = S_r - S_{r-1}$?

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  • $\begingroup$ Hint: simple trick in this case: $\;v_{n+1}:=u_{n+1}+r=4(u_n +r)=4\,v_n\;$ so... $\endgroup$ Feb 6 '20 at 7:53
  • $\begingroup$ Could you please explain how to get the r? Thanks $\endgroup$
    – gc3941d
    Feb 6 '20 at 7:58
  • $\begingroup$ Well, what is $\,u_{n+1}-4\,u_n$ in both cases? $\endgroup$ Feb 6 '20 at 9:38
  • $\begingroup$ Hi @RaymondManzoni I think I haven’t learned $v_{n+1} = u_{n+1} + r$ ...is it a formula? $\endgroup$
    – gc3941d
    Feb 6 '20 at 15:45
  • $\begingroup$ In fact we define (that's the meaning of $:=$) $v_n$ to be $\,v_n:=u_n+r\,$ (for all positive $n$ and thus $\,v_{n+1} = u_{n+1} + r$) with the idea that $v_n$ will be easier to handle than $u_n$. $\endgroup$ Feb 6 '20 at 17:33
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$u_2=4u_1+7$

$u_3=4(4u_1+7) + 7 = 16u_1+ 4\times 7 + 7$

$u_4= 4^3u_1 +7(4^2+4+1)$

$u_r=4^{r-1}u_1 + 7(4^{r-2}+4^{r-3}.......+1)$ (after finding the rth term by observing the pattern then you can use induction to prove it)

$u_r=4^{r-1} + 7(\frac{4^{r-1}-1}{3})\ =\ 1+\frac{10}{3}(4^{r-1} - 1)$

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  • $\begingroup$ I get the pattern but I don’t get the last line. Why is there a fraction with denominator 3? And how come it’s $4^{r-1}$ minus 1? Thanks @aryan bansal $\endgroup$
    – gc3941d
    Feb 7 '20 at 1:42
  • $\begingroup$ do you mean $S=\frac {a}{1-r} ?$ $\endgroup$
    – gc3941d
    Feb 7 '20 at 4:08
  • $\begingroup$ That's for infinite terms, i am talking about finite terms G.P, it's $\frac{(a)(r^n-1)}{(r-1)}$ where '$n$' is number of terms, '$r$' is common ratio and '$a$' is the first term. $\endgroup$ Feb 7 '20 at 4:12
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for n=1 is ok, let's take it true for n and prove it for n+1. $u_{n+1}=4u_n+7=4+40/3[4^{r-1}-1]+7=11+10/3[4^r-4]=1+30/3-40/3+10/3[4^r]=1+10/3[4^r-1]$

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  • $\begingroup$ Sadly, this will not suffice for this question. It's clearly an examination question, the "Show that" implies just verifying the formula is insufficient, you must derive the formula from the recurrence relation. $\endgroup$ Feb 6 '20 at 8:06
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Let $u_n=4^nv_n$.

Now you have

$$4^{n+1}v_{n+1}=4v_n4^n+7$$ or

$$v_{n+1}=v_n+\frac7{4^{n+1}}.$$

So by induction

$$v_n=v_1+\frac74\sum_{k=1}^n\frac1{4^k}.$$

The rest is yours.

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