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

Question

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 $u_r = S_r - S_{r-1}$?

Answer 1

$u_2=4u_1+7$

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

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

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

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

Answer 2

for n=1 is ok, let's take it true for n and prove it for n+1. $u_{n+1}=4u_n+7=4+40/3[4^{r-1}-1]+7=11+10/3[4^r-4]=1+30/3-40/3+10/3[4^r]=1+10/3[4^r-1]$

Answer 3

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.