Question

what is the rule for the nth turn
example a sub n =n
for the set 1, 6, 11, 16, 21...

Answer 1

the example was
\[a _{n}=n\]

Answer 2

@dan815 @zepdrix

Answer 3

Hmm are you able to see a pattern in your numbers? :)
Are we adding something each time?
Multiplying by something each time?

What type of change is happening from 1 to 6,
from 6 to 11,
and so on?

Answer 4

adding 5

Answer 5

that i knew the others i did not

Answer 6

plus 5! Ok great.
Let's come up with a rule for the second term,
\(\large\rm a_2=a_1+5\)
yes?

How bout the third term,
we would add 5 to the first term, twice,
\(\large\rm a_3=1+5+5\)
or we can write it this way,
\(\large\rm a_3=1+5(2)\)

Answer 7

but what about the first

Answer 8

Well, we start at the first term,
so we're adding none of those 5's, right?

Answer 9

\[\large\rm a_1=1+5(0)\]

Answer 10

so how about
\[a _{n}=1+5(n-1)\]

Answer 11

Mmm ya I like that idea! :)
Maybe you noticed that the number of 5's we need is always `one less than our term`.\[\large\rm a_{\color{orangered}{2}}=1+5(\color{orangered}{1})\]\[\large\rm a_{\color{orangered}{3}}=1+5(\color{orangered}{2})\]\[\large\rm a_{\color{orangered}{4}}=1+5(\color{orangered}{3})\]\[\large\rm a_{\color{orangered}{n}}=1+5(\color{orangered}{n-1})\]So yes :) gj

Answer 12

what if the set was 1, 2, 4, 8, 16 you would multiply by 2

Answer 13

@phi

Answer 14

@zepdrix

Answer 15

yes, you multiply the "previous value" to get the next value

Answer 16

what is the rule

Answer 17

or , for n (the term) =1,2,3,4,
the corresponding value is 2^(n-1)

Answer 18

thanks

Answer 19

notice (ignoring the 1 for the moment)
your numbers
1,2,4,8,16
can be written
1, 2, 2*2, 2*2*2, 2*2*2*2

Answer 20

if you know 2^0 is 1
and 2 is the same as 2^1
and 2*2 is 2^2
you see the pattern
2^0, 2^1, 2^2, 2^3, 2^4

and if we count the terms using n=1,2,3,4,
the nth term is 2^(n-1)

Answer 21

For example, the first term (when n=1) gives the answer
2^(1-1) = 2^0 = 1

Answer 22

true
what if the signs swap like -4, 8, -12, 16, -20 what happens to the rule because you have to multiply and add
sorry for asking to much

Answer 23

one way is to first ignore the signs, so you have
4,8,12,16,20

can you write a formula for then nth term ?

Answer 24

\[a _{n}=4n\]

Answer 25

if you use the formula 4n with n=1,2,3,etc
we get
4, 8, 12, 16, etc
that looks good

to get alternating minus signs we use a trick
(-1)^n when n is odd gives us -1
(-1)^n when n is even gives us +1
so multiply your formula 4n by (-1)^n

(-1)^n 4n

Answer 26

\[a _{n}=(-1)^{4n}\]

Answer 27

\[ a_n= (-1)^n 4 n \]

Answer 28

got it

Answer 29

some are though unfamiliar those were easy to pop out

Answer 30

this one is unfamiliar how do you do that there is niether a common difference or common ratio
2, 9, 28, 65, n=5

Answer 31

does the n=5 mean they want the 5th term?

Answer 32

need rule

Answer 33

and 5th

Answer 34

you can find differences of differences
like this
2, 9, 28, 65
7 19 37
12 18
6

we have to assume that the 6 represents the pattern
thus, to extend the pattern we do

2, 9, 28, 65
7 19 37
12 18
6 6

the next row above the 6 will be 18+6

2, 9, 28, 65
7 19 37
12 18 24
6 6

extending the row above the 24, we get 37+24

2, 9, 28, 65
7 19 37 61
12 18 24
6 6

finally, the top row is extended by 65+61
2, 9, 28, 65 126
7 19 37 61
12 18 24
6 6

Answer 35

desrcibe and write the rule

Answer 36

its ++5 every time so

Answer 37

after 5 times it will be plus n*5

Answer 38

no the new one not the 5

Answer 39

2 9 28 65 i need n=5 and rule

Answer 40

http://prntscr.com/a7kgai

Answer 41

i dont get the rule though

Answer 42

i dont get the rule though