1. ## Sequences/Patters

given the patterns
a) 4,8,12,16,20,24(going up by 4)
b) 3,9,27,81,243,729(* by 3)
c) 144,121,100,81,64,49(-21,-21,-19...)
d) 3,-12,48,-192,728,-3072(*-4)

Find an expression for the general term for the sequences listed above(one general term for each sequence)
Can you please explain how to do it aswell
thanks

2. [quote=euclid2;186568]given the patterns
a) 4,8,12,16,20,24(going up by 4)

$\displaystyle 4,8,12,16,20............ \hfill \\$

$\displaystyle {\text{This arithmetic sequence has first term, }}a = 4{\text{ and common diffrence, }}d = 8 - 4 = 4 \hfill \\$

$\displaystyle {\text{General term, }}t_n = a + \left( {n - 1} \right)d \hfill \\$

$\displaystyle = 4 + \left( {n - 1} \right)4 \hfill \\$

$\displaystyle = 4 + 4n - 4 = 4n \hfill \\$

$\displaystyle t_n = 4n \hfill \\$

3. [quote=euclid2;186568]given the patterns

b) 3,9,27,81,243,729(* by 3)

$\displaystyle 3,9,27,81,243,................... \hfill \\$

$\displaystyle {\text{This geometric sequence has first term, }}a = 3{\text{ and common ratio, }}r = \frac{9} {3} = 3 \hfill \\$

$\displaystyle {\text{General term, }}t_n = ar^{n - 1} \hfill \\$

$\displaystyle = 3\left( 3 \right)^{n - 1} \hfill \\$

$\displaystyle = 3^{1 + n - 1} = 3^n \hfill \\$

$\displaystyle t_n = 3^n \hfill \\$

4. [quote=euclid2;186568]given the patterns

d) 3,-12,48,-192,728,-3072(*-4)

$\displaystyle 3, - 12,48, - 192,728, - 3072,................... \hfill \\$

$\displaystyle {\text{This geometric sequence has first term, }}a = 3{\text{ and common ratio, }}r = \frac{{ - 12}} {3} = - 4 \hfill \\$

$\displaystyle {\text{General term, }}t_n = ar^{n - 1} \hfill \\$

$\displaystyle = 3\left( { - 4} \right)^{n - 1} \hfill \\$

$\displaystyle t_n = 3\left( { - 4} \right)^{n - 1} \hfill \\$

5. thank you very much that is extremely helpful

6. [quote=euclid2;186568]given the patterns

c) 144,121,100,81,64,49(-21,-21,-19...)

$\displaystyle 144,121,100,81,64,49,................. \hfill \\$

$\displaystyle 12^2, 11^2, 10^2, 9^2,...............$

$\displaystyle {\text{This sequence is neither arithmetic nor geometric}} \hfill \\$

$\displaystyle \text { the sequence is}\;\; (13-1)^2, (13-2)^2, (13-3)^2,......$

$\displaystyle t_n=(13-n)^2$

7. Hello, euclid2!

Given the patterns:

a) 4, 8, 12, 16, 20, 24, ...

b) 3, 9, 27, 81, 243, 729, ...

c) 144, 121, 100, 81, 64, 49, ...

d) 3, -12, 48, -192, 728, -3072, ...

Find an expression for the general term for the sequences.

We must examine the pattern and create a function in terms of $\displaystyle n.$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle a)\quad\begin{array}{c|c|c|c|c|c|c| c} n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline a_n & 4 & 8 & 12 & 16 & 20 & 24 & \hdots \\ & 4\cdot{\color{red}1} & 4\cdot{\color{red}2} & 4\cdot{\color{red}3} & 4\cdot{\color{red}4} & 4\cdot{\color{red}5} & 4\cdot{\color{red}6} & \hdots \\ \hline \end{array}$

We see that $\displaystyle a_n$ is always: $\displaystyle \text{4 times }n.$

Therefore: . $\displaystyle a_n \;=\;4n$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle b)\quad\begin{array}{c|c|c|c|c|c|c|c} n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline a_n & 3 & 9 & 27 & 81 & 243 & 729 & \hdots \\ & 3^{{\color{red}1}} & 3^{{\color{red}2}} & 3^{{\color{red}3}} & 3^{{\color{red}4}} & 3^{{\color{red}5}} & 3^{{\color{red}6}} & \hdots \\ \hline \end{array}$

We see that $\displaystyle a_n$ is always: $\displaystyle \text{3 to the }n^{th}\text{ power.}$

Therefore: . $\displaystyle a_n \;=\;3^n$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle c)\quad\begin{array}{c|c|c|c|c|c|c|c} n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline a_n & 144 & 121 & 100 & 81 & 64 & 49 & \hdots \\ & 12^2 & 11^2 & 10^2 & 9^2 & 8^2 & 7^2 & \hdots \\ & (13-{\color{red}1})^2 & (13-{\color{red}2})^2 & (13-{\color{red}3})^2 & (13-{\color{red}4})^2 & (13-{\color{red}5})^2 & (13-{\color{red}6})^2 & \hdots \end{array}$

We see that $\displaystyle a_n$ is always: $\displaystyle \text{the square of }(13-n)$

Therefore: . $\displaystyle a_n \;=\;(13-n)^2$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle d)\quad\begin{array}{c|c|c|c|c|c|c|c} n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline a_n & 3 & \text{-}12 & 48 & \text{-}192 & 728 & \text{-}3072 & \hdots \\ & 3(\text{-}4)^{{\color{red}0}} & 3(\text{-}4)^{{\color{red}1}} & 3(\text{-}4)^{{\color{red}2}} & 3(\text{-}4)^{{\color{red}3}} & 3(\text{-}4)^{{\color{red}4}} & 3(\text{-}4)^{{\color{red}5}} & \hdots\end{array}$

We see that $\displaystyle a_n$ is always: $\displaystyle \text{3 times a power of -4,}$
. . $\displaystyle \text{and that power is }n-1.$

Therefore: . $\displaystyle a_n \;=\;3\cdot(\text{-}4)^{n-1}$