help please (Nth term in maths)?

Question

i got to work out the sequences Nth term can some one help the sequence is 2,5,10,17,26........nd the 2nd one is 2, 6, 12, 20, 30, ........thanks for your help

Answer 1

n^2 + 1
n^2 + n

Answer 2

1. The differences between these are 3,5,7,9...
The differences between these then are all 2
therefore nth term=n(n+1) or n squared +1
check 1st term = 1(1+1)=2,
2nd term = 2(2+1)=5, etc

2. The differences between these are 4, 6, 8, 10
The differences between these again are all 2
therefore nth term=n squared +n
check 1st term = 1 squared +1 =2
2nd term = 2 squared +2 =6, etc

Hope you find this helpful :)

Answer 3

In the first case, the nth term is n^2+1 and in the second case, the nth term is n(n+1)...

SOURCE: http://www.mathsrevision.net/gcse/pages.php?page=9

Answer 4

1st series

The differences between consecutive numbers are 3, 5, 7...
The differences between the differences are consistently 2.
This implies the series involves squares

x(n) = n² + c

By observation c = 1

nth term is n² + 1
----------------------------
2nd series

Differences are 4, 6, 8, 10
Again the differences of the differences are 2, so it involves squares

n² + c

x(1) = 1 + c --> c = 1
x(2) = 4 + c --> c = 2
x(3) = 9 + c --> c = 3

By observation c = n

nth term is n² + n

Answer 5

1st sequence: the nth term is n^2 + 1
2nd sequence: the nth term is n(n+1)

To get these, you can just look for a pattern, or look at the differences between consecutive terms, and if necessary, the differences between consecutive differences.

Answer 6

1)1st sequence
2,5,10,17,26
1st difference
3,5,7,9
2nd difference
2,2,2
this is a polynomial of the
second order
N=an^2+bn+c....(1)
n=1,2,3.........
draw up a system of equations
using the first three terms
2=a+b+c
5=4a+2b+c
10=9a+3b+c
1st sweep
3=3a+b
2nd sweep
5=5a+b
2=2a,a=1
sub back,b=0,c=1
substitute these values into (1)
N=n^2+1, n=1,2,3,........

1)2nd sequence
2,6,12,20,30
1st difference
4,6,8,10
2nd difference
2,2,2
this is a polynomial of the
second order
N=an^2+bn+c....(2)
n=1,2,3.........
draw up a system of equations
using the first three terms
2=a+b+c
6=4a+2b+c
12=9a+3b+c
1st sweep
4=3a+b
2nd sweep
6=5a+b
2=2a,a=1
sub back,b=1,c=0
substitute these values into (2)
N=n^2+n,
=n(n+1), n=1,2,3,........

i hope that this helps

Answer 7

The first one is
n squared plus 1
the second is
n squared plus n

no problem

Answer 8

the first one, 2, 5,... can be seen to be summing up odd numbers, starting from base 0, that is
F(1)=2
F(2)=5=F(1)+3
F(3)=10=F(2)+5
F(n)=F(n-1)+2n-1

The second one 2, 6, is similar except is sums evens
G(1)=2
G(2)=6=G(1)+4
G(3)=12=G(2)+6
G(n)=G(n-1)+2n

Answer 9

i think its n+2 and n+4 i hent done nth term for a while but thats what i remember