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 $u_r = S_r – S_{r-1}$?
Best Answer
$u_2=4u_1+7$
$u_3=4(4u_1+7) + 7 = 16u_1+ 4\times 7 + 7$
$u_4= 4^3u_1 +7(4^2+4+1)$
$u_r=4^{r-1}u_1 + 7(4^{r-2}+4^{r-3}.......+1)$ (after finding the rth term by observing the pattern then you can use induction to prove it)
$u_r=4^{r-1} + 7(\frac{4^{r-1}-1}{3})\ =\ 1+\frac{10}{3}(4^{r-1} - 1)$