Question

find the nth term
1 2 3 4 5 6
2 8 18 32 50 ?
show work!

Answer 1

which nth term :)

Answer 2

6

Answer 3

1=2, 2=8, 3=18, 4=31 5=50 6=?

Answer 4

2 = 2
2+6 = 8
8+10 = 18
18+14 = 32

not a pattern there

Answer 5

got it

Answer 6

n = (n-1)+1; n1 = 1

Answer 7

an + 2(1,3,5,7,9,...)

Answer 8

Terms : 2 8 18 32 50 72
Diffs : 6 10 14 18 22
2nd diffs : 4 4 4 4

Answer 9

well, if we start at:
a1 = 0 + 2(1) = 2
a2 = 2 + 2(3) = 8
a3 = 8 + 2(5) = 18
a4 = 18 + 2(7) = 32

Answer 10

so whats the nth term?????

Answer 11

a5 = 32 + 2(9)
32 + 18
50 woohoo!!!

Answer 12

do you want the 6th or the nth ?

Answer 13

nth

Answer 14

the 6th is 72

Answer 15

joe is always really good at these

Answer 16

an = 2(2n-1) maybe :)

Answer 17

These problems dont make sense to me anymore =/

Answer 18

lol

Answer 19

but whats the nth!

Answer 20

There is an algorithm for this . Just answered a similar Q about it - hold on.

Answer 21

an + 2(2n-1)

since the odds are 2k -1

Answer 22

We can use the formula

tn = a+(n-1)d
Where a = first number in the equation
n = term number
d = common difference

Solve for n

Answer 23

almost got my brain around it lol

Answer 24

to follow what the person who make the problem wants us to do, since the 2nd differences are constantly 4, its a quadratic polynomial in the form an^2+bn+c

Answer 25

oh whats the fun in using premade wheels ... i like to reinvent them as i go ;)

Answer 26

you can plug in values of n (1, 2, 3,...) and create a system of equations.

Answer 27

n(2n)
or
2n^2

Answer 28

so we see
2=2(1)
8=2(2(2))
18=3(2(3))
32=4(2(4))
so on....

Answer 29

an = an-1 + 2(2n-1)
= an-2 + 2(2n-1) + 2(2n-1)
= an-3 + 2(2n-1) + 2(2n-1)+ 2(2n-1)

myin seems to be onto something easier lol

Answer 30

for the record :)
http://www.mathsisfun.com/algebra/sequences-finding-rule.html

Answer 31

so the nth term is an-3+2(2n-1)+2(2n-1)+2(2n-1)???? and thanks! haha

Answer 32

i wish there was set "rules" :)

Answer 33

its just 2n^2.

Answer 34

it prolly simplifies out to myins

Answer 35

a1 = 2.1.1
a2 = 2.2.2
a3 = 2.3.3
an = 2.n.n = 2n^2

Answer 36

divide those terms by 2, they are perfect squares. thats probably what the creator of the problem intended.

There are infinitely many formulas that could work though, and infinitely many "6th terms"

Answer 37

And Joe is right :)
http://www.wolframalpha.com/input/?i=2+++++8++++++18++++++32+++++50

Answer 38

i seen n(2n) and then the simplification or product is 2n^2
way to multiply joe!
thats why you are one of my favorites ;)

Answer 39

what would you do-o-o 4 a joemath bar lol

Answer 40

lol <.<

Answer 41

this link will work:
http://www.wolframalpha.com/input/?i=2+8++18++32++50

Answer 42

i want a heath bar with some vanilla icecream

Answer 43

the links are on the fritz by the way

Answer 44

give me 5 numbers between -5 and 5, without picking 0.

Answer 45

.031172 is my first pick

Answer 46

hahah thanks everyone!

Answer 47

i guess i should specify integers, to make the calculations easier on myself lol

Answer 48

;)

Answer 49

1 2 3 4 5

Answer 50

3 1 1 4 2 are 5 nukbers

Answer 51

...the k key is my new nemsis

Answer 52

lol

Answer 53

alright, give me a couple of minutes, i'll pop out another formula that fits those first 5 numbers for n = 1, 2, 3, 4, 5, and gives some other 6th term.

Answer 54

"between -5 and 5 tends to mean up to but not including ... mathically that is

Answer 55

http://www.wolframalpha.com/input/?i=-14%2F3%281%29^n%2B10%2F3%282%29^n%2B1%2F3%283%29^n-1%2F3%284%29^n%2B1%2F15%285%29^n%2C+n+%3D+1

you can change the value of n at the end of the equation to verify. According to this formula, the 6th term is 128 lol

Answer 56

im actually thinking about writing an undergraduate thesis on this idea. <.<