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nth term for sequences

4

hello i need to find a formula for the following sequnce :

(please note that they are two one's are the beginning that isn't a mistake)

1,1,7,25,63,129,231

i have tried so many times but still i ain't getting the right answer when i try out the formula.

can anyone please help me out.

thanks :rolleyes:

Reply 1

Chewwy

17

disregarding the first term, (2n+1)(2n^2+2n+3)/3 works...

if you replace all the ns with (n-1)s, the mod will give the right sequence. i.e.

|(2(n-1)+1)(2(n-1)^2+2(n-1)+3)/3| = |(2n-1)(2n^2 - 4n + 2 + 2n - 2 + 3)/3| = |(2n-1)(2n^2 - 2n + 3)/3|

Reply 2

*Sarah...*

Chewwy, did you have a method/formula for finding that or was it simply trial and error?

Reply 3

Chewwy

17

*Sarah...*

Chewwy, did you have a method/formula for finding that or was it simply trial and error?

oh, i had a method. i looked here - http://www.research.att.com/~njas/sequences/index_b.html

:biggrin:

Reply 4

*Sarah...*

chewwy

oh, i had a method. i looked here - http://www.research.att.com/~njas/sequences/index_b.html

:biggrin:

lol fair enough... :p:

Reply 5

rpatel2

OP

4

is there a formula when you start of with the two one's

Reply 6

*Sarah...*

rpatel2

is there a formula when you start of with the two one's

i was thinking along the lines of (1)^n... but can't think what the rest could be :confused:

Reply 7

damaster275

1

If you look at the successive common diferences, then u can see that the sequence is a cubic. and u no that a cubic is of the form ax^3+bx^2+cx+d and thus u can form equations for each term of the sequence and solve for the constants a,b,c and d. Hope that helps :biggrin:

Reply 8

steve2005

17

damaster275

If you look at the successive common diferences, then u can see that the sequence is a cubic. and u no that a cubic is of the form ax^3+bx^2+cx+d and thus u can form equations for each term of the sequence and solve for the constants a,b,c and d. Hope that helps :biggrin:

I have tried the cubic and it does not give the correct sequence. Perhaps I have made a mistake.. but I don't think so.

Picture 1.png87.5KB

Reply 9

damaster275

1

steve2005

I have tried the cubic and it does not give the correct sequence. Perhaps I have made a mistake.. but I don't think so.

Lol, erm just ignore me then.

Actually, looking at the third difference which is 8, the coeefficient of the x^3 term should be 8/6 or 4/3 but ur calculation seems immaculate. Strange....

Reply 10

rpatel2

OP

4

it seems to me that the 1 in the beginning of the sequence is mis typed number. from now and on conisder the sequence as : 1,7,25,63,129,231

by any chance can you use sigma notation to calculate these? if so how is it done?

Reply 11

carol05

rpatel2

1,7,25,63,129,231

(4n³-6n²+8n-3)/3

Reply 12

damaster275

1

rpatel2

it seems to me that the 1 in the beginning of the sequence is mis typed number. from now and on conisder the sequence as : 1,7,25,63,129,231

by any chance can you use sigma notation to calculate these? if so how is it done?

Lol use my method.

Reply 13

*Sarah...*

carol05

(4n³-6n²+8n-3)/3

how did you get that?

Reply 14

carol05

*Sarah...*

how did you get that?

I used steve's method
a+b+c+d=1 (1)
8a+4b+2c+d=7 (2)
(2)-(1) 7a+3b+c=6 (3)
27a+9b+3c+d=25 (4)
(4)-(1) 26a+8b+2c=24 (5)
2 x (3) 14a +6b +2c=12 (6)
(5)-(6) 12a +2b =12
6a +b = 6 (7)
64a+16b+4c+d=63 (8)
(8)-(1) 63a+15b+3c = 62 (9)
21a +5b +c = 62/3 (10)
(10)-(3) 14a+2b=62/3 -6 (11)
2 x (7) 12a +2b =12
2a= 20.3333- 18
a= 1.333
b=6-6(1.333)=-2
c=2.6666
d= -1
1.333n³-2n²+2.666n-1 = (4n³-6n²+8n-3)/3

Reply 15

mathsexams

You can easily find the nth term of *any* finite sequence using the formula:

a + (n-1)b + (n-1)(n-2)c/2! + (n-1)(n-2)(n-3)d/3! + (n-1)(n-2)(n-3)(n-4)e/4! + ...

where a=1st term and b,c,d,e,etc are the first number in the first, second, third etc difference rows.

a=1 1 7 25 63 129 231 sequence
b=0 6 18 38 66 102 1st diffs
c=6 12 20 28 36 2nd diffs
d=6 8 10 12 3rd diffs
e=2 2 2 4th diffs
f=0 0 5th diffs (stop when 0)

which gives:

nth term = 1 + 3(n-1)(n-2) + (n-1)(n-2)(n-3) + (n-1)(n-2)(n-3)(n-4)/12
[edit: forgot (n-4)]

which you can multiply out if you want :smile:

Reply 16

steve2005

17

rpatel2

it seems to me that the 1 in the beginning of the sequence is mis typed number. from now and on conisder the sequence as : 1,7,25,63,129,231

by any chance can you use sigma notation to calculate these? if so how is it done?

I don't know how you would use sigma for sequences... but this might help
http://planetmath.org/encyclopedia/Summing.html

http://home.alltel.net/okrebs/page136.html
http://www.mathcentre.ac.uk/students.php/all_subjects/series/sigma/resources/

The pic may be what your teacher wants. ( I am only guessing)

Picture 4.png6.9KB

Reply 17

reckless

3

mathsexams

You can easily find the nth term of *any* finite sequence using the formula:

a + (n-1)b + (n-1)(n-2)c/2! + (n-1)(n-2)(n-3)d/3! + (n-1)(n-2)(n-3)(n-4)e/4! + ...

where a=1st term and b,c,d,e,etc are the first number in the first, second, third etc difference rows.

a=1 1 7 25 63 129 231 sequence
b=0 6 18 38 66 102 1st diffs
c=6 12 20 28 36 2nd diffs
d=6 8 10 12 3rd diffs
e=2 2 2 4th diffs
f=0 0 5th diffs (stop when 0)

which gives:

nth term = 1 + 3(n-1)(n-2) + (n-1)(n-2)(n-3) + (n-1)(n-2)(n-3)/12

which you can multiply out if you want :smile:

You know what: I did exactly the same thing yesterday after seeing this question and multiplied it out - not very nice, and I made a mistake somewhere so ended up with the wrong polynomial and couldn't be bothered to correct it and post it. - Too much work.

Reply 18

Chewwy

17

steve2005

The pic may be what your teacher wants. ( I am only guessing)

wha!?!?

nth term for sequences

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