A student wrote that a number pattern such that each number is 3 more than 4 times the previous number. What is the sixth term in the pattern if the first term is 6?

Question

jhannybean · Answer

$$3+4(n-1) $$ where $$a_1 = 6$$

jack1 · Answer

the 6th term in the series using that formula would be 23... assuming n=a1=6... yeah?

jack1 · Answer

is that right?

jhannybean · Answer

n represents many numbers, so if we take the first number, 6, you would just plug that in, i believe

jack1 · Answer

im not sure that's right... the 2nd term should b 27 in the series, if the first term is 6

jhannybean · Answer

How would the equation be written for the sequence?

hunus · Answer

http://labs.codecademy.com/BBcN#:workspace Here's a cool little program that will show you the numbers in the sequence

hunus · Answer

$$\Large a_{n}=4a_{n-1} + 3$$
Where 
$$a_1 = 6$$

jhannybean · Answer

Oh, you're right, Jack1. Since this is a pattern of adding +4 to every number in the sequence, we consider this an arithmetic sequence. 
Hunus wrote the equation out better :P

jack1 · Answer

1st term 6
2nd term 27
3rd term 111
4th term 447
... have i got the right pattern here...?

hunus · Answer

Yup, look at the link I posted

jhannybean · Answer

I'm not entirely sure what to input in there, and it's a bit confusing :P

hunus · Answer

Ah, just press the run button at the top and it will ask you which term you want it to compute up to

jack1 · Answer

@Hunus can you explain your equation there?
is that to the power of anything?

hunus · Answer

Nope, it says the nth term of a sequence$$\Large a_n$$is three plus the term before that one $$\Large a_{n-1}$$ multiplied by four

jack1 · Answer

ok, so how does your equation take into account the previous term...?
sorry, maybe i'm reading it wrong, but to me that equation looks like:

a(n)   =   [4   x   a   x   (n-1) ]  +3

...is that right?

jack1 · Answer

and a = 6

hunus · Answer

The previous term is a_(n-1)

hunus · Answer

a_1 = 6
a_2 = 4*a_1 + 3 = 4*6+3

jack1 · Answer

but that only lets you compute them sequentially, as you need the previous term

jack1 · Answer

it's not really an equation that lets you work out any term of a series

jack1 · Answer

yeah...? or am i still not understanding?

hunus · Answer

You're correct, you would have to do them one by one

jack1 · Answer

ok, sorry it wasn't a put down, i'm here for the knowledge, and i thought I was missing something

jack1 · Answer

what about for the whole series:
term (n) in the series = ( 1st term in the series x 4^(n-1) ) + ( (4^(n-1) -1)

$$a _{n} = \left\{ a _{1} \times4^{(n-1)}\right\} + \left[ 4^{(n-1)} -1\right]$$

jack1 · Answer

so the first 6 terms are 6, 27, 111, 447, 1791, 7167... and now all you have to do is plug in 
a1 = 6
and which term you want (n)...

hunus · Answer

Yup