It is easy to find a rule based on a polynomial of order 4 such that the first four numbers are as listed in the question. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.
One possible answer is t(n) = 3n - 8
another is (n^4 - 10*n^3 + 35*n^2 - 26*n - 40)/8
still another is (n^4 - 10*n^3 + 35*n^2 - 41*n)/3
Assuming each term is 3 MORE than the previous term t(n) = -13 + 3*n where n = 1, 2, 3, ...
The formula is Un = n*(-1)n+1
The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.
a(n) = a(1) x r^n-1 In this case a(1) is 2 and r is 3 That makes the formula a(n) = 2 x 3^n-1
It looks as if each number is equal to the previous number, multiplied by -1/2. Therefore, one way to write the formula for the n'th term is:-24 x (-1/2)^n
The answer depends on what the explicit rule is!
The simplest formula isUn = (-8611*n^2 + 34477*n - 25082)/2 for n = 1, 2, 3.
The explicit formula for a sequence is a formula that allows you to find the nth term of the sequence directly without having to find all the preceding terms. To find the explicit formula for a sequence, you need to identify the pattern or rule that governs the sequence. This can involve looking at the differences between consecutive terms, the ratios of consecutive terms, or any other mathematical relationship that exists within the sequence. Once you have identified the pattern, you can use it to create a formula that will generate any term in the sequence based on its position (n) in the sequence.
Assuming each term is 3 MORE than the previous term t(n) = -13 + 3*n where n = 1, 2, 3, ...
An explicit formula is a formula in which depicts relations between the sums over complex number zeros and over prime numbers. An example of an explicit formula is: _(t) = _log(_) + Re(_(1/4 + it/2)).
a recursive formula is always based on a preceding value and uses A n-1 and the formula must have a start point (an A1) also known as a seed value. unlike recursion, explicit forms can stand alone and you can put any value into the "n" and one answer does not depend on the answer before it. we assume the "n" starts with 1 then 2 then 3 and so on arithmetic sequence: an = a1 + d(n-1) this does not depend on a previous value
Each term is 3 times greater than the previous term and so the next term will be 486
Each term is the sum of the 2 preceding terms; where the first 2 terms are 1 and 1. So 1, 1, 2, 3, 5, 8, 13, 21 etc.
arithmetic sequence * * * * * A recursive formula can produce arithmetic, geometric or other sequences. For example, for n = 1, 2, 3, ...: u0 = 2, un = un-1 + 5 is an arithmetic sequence. u0 = 2, un = un-1 * 5 is a geometric sequence. u0 = 0, un = un-1 + n is the sequence of triangular numbers. u0 = 0, un = un-1 + n(n+1)/2 is the sequence of perfect squares. u0 = 1, u1 = 1, un+1 = un-1 + un is the Fibonacci sequence.
The formula is Un = n*(-1)n+1
The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.
a(n) = a(1) x r^n-1 In this case a(1) is 2 and r is 3 That makes the formula a(n) = 2 x 3^n-1