non-linear sequences?

Question

how do ya find the nth term??

erm, aslo if you had the seqeuence 10 divided by 10, will it be 1, 10, 100, 1000.....????
i love u all for answering my dears...

Answer 1

This is a type of question that appears in tests quite frequently. There are probably an infinite set of non-linear sequences, so there is no formula for calculating the nth term. On tests, you will have to examine all the numbers relative to each other, in order, to figure out what the sequence represents.

Simple example:
1, 4, 9, 16, .... is the sequence of intergers squared.

Harder example:
1, 100, 1001, 10000, .... is the binary sequence of decimal integers squared.

Impossible example:
1, 0, 2, 4 ...is the number of scars on the fingers of my right hand!

Hope this helps!

Answer 2

To find the nth term follow these steps:

1)Find the gap between each number to the previous number

If this is equal you have a linear equation eg n+2 . General equation is n+c where c is a constant term

2)If the gaps are not equal, then take the gaps of these numbers.

If they are equal now you have a quadratic equation
eg n^2+2. General is n^2+c where c is constant term

Example

The sequence 2,5,10,17,26,37

Take the gaps gives u: 5-2,10-5,17-10,26-17,37-26
Which is 3,5,7,9,11

The gaps are not equal so you take the gaps again, which is
2,2,2,2

The gaps are now equal so you know you have a quadratic of
n^2 + c

The 1st term of the sequence means n=1 which is 2

So to get from n=1 to 2 you do x^2+c =2
1+c= 2
c=1

So your equation is n^2+1

For gaps which aren't equal after taking gaps twice, you just continue in the same mannar

For your additional information the sequence 10/10 will be:
1,1,1,1,1,1,...

For the sequence to be 1,10,100,1000
the nth term will be (10^n)/10 or 10^(n-1)

A sequence must include n as 10/10 does not include n the sequence will continue to be 10/10 = 1 forever.

Hope it helps a lot
OZ

Answer 3

it is not easy -hit the site below to get an insight into
some of the easier sequences-you don't want to
spend your life trying to break down the
impossible!

http://www.physicsforums.com/showthread.php?t=114995

i'll give a non-rigorous proof of how it works

here N the nth term described by
n=1 instead of the usual n=1,2,3,.............

we have the sequence described by
N=(n+1)(n+2)say,a second order
1st term
N=n^2+3n+3
2nd term-add 1 to n
N=n^2+5n+6
3rd term-add 1 to (n+1)
N=n^2+7n+12
4th term
N=n^2+9n+20
5th term
N=n^2+11n+30

1st diff
2n+4
2n+6
2n+8
2n+10
notice how the first difference sweeps away the
second order component of the sequence
description
2nd diff
2,
2,
2,
notice how the second difference sweeps away
the first order component of the sequence
description and you are left with a
constant
that's ok with polynomials-it becomes more
complicated when you go into exponents

i hope that this helps

Answer 4

First show f(n) is true; then show f(n+1) - f(n) is true. If both are true then you have found your sequence in terms of n.

Example f(x^n) = x^1 + x^2 + ... + x^n
and f(x^(n+1)) = x^1 + x^2 + ... + x^n + x^(n+1)

So that f(x^(n+1) = f(x^n) + x^(n+1) and f(x^(n+1)) - f(x^n) = x^(n+1) is true; so x^n is a proper nonlinear sequence in n.

Answer 5

Calculate it according to the definition of the sequence.