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It is impossible to "solve" the formula since you will be given only a finite number of values. If you are given k values then there is a polynomial of order (k-1) that will generate those values, and infinitely more polynomials of higher order which will do so. Furthermore, there are non-polynomial functions that will do the trick as well.

Having said that, there are some things you can do towards solving the formula. The question is usually answered using Occam's razor: if there are two or more possible solutions, use the simpler one.


If the question mentions arithmetic sequence, then you know that each term in the sequence is equal to the preceding term plus some constant (which may be negative). This is known as the "common difference".


The position to term formula for an arithmetic sequence is:

U(n) = a + d*n for the nth term,

where n is a counter that locates the term in the sequence (n = 1, 2, 3, ...)

d is the common difference and

a is the 0th term. That is, the term that would have come before the first term if you continued the sequence for one step in the reverse direction.


There are polynomial sequences, where the first round of calculating differences between successive terms does not yield a constant but differencing the sequence formed by these differences (the second difference) is a constant. In this case the solution is a quadratic rule. Similarly, if the third differences are the same, the rule is cubic and so on.


If the question mentions geometric sequence then that shows that each term is a fixed multiple (which may be smaller than 1, or negative) of the preceding term. This is known as the "common ratio".


The position to term formula for a geometric sequence is:

U(n) = a + r^n for the nth term,

where n is a counter that locates the term in the sequence (n = 1, 2, 3, ...)

r is the common difference and

a is the 0th term. That is, the term that would have come before the first term if you continued the sequence for one step in the reverse direction.


Then there are special sequences that students are often expected to recognise. These include:

1, 3, 6, 10, 15, ... (triangular numbers - the second differences are a constant)

1, 4, 9, 16, 25, ... (square numbers - the second differences are a constant)

2, 3, 5, 7, 11, 13, ... (prime numbers)

1, 1, 2, 3, 5, 8, 13, ... (the Fibonacci sequence - defined by U(1) = 1, U(2) = 1 and U(n) = U(n-2)+U(n-1) for all n >2.)

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Q: How do you solve the formula of a sequence of numbers.?
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