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
This is a sequence based on the squares of numbers (positive integers) but starting with the square of 2. Under normal circumstances the sequence formula would be n2 but as the first term is 4, the sequence formula becomes, (n + 1)2. Check : the third term is (3 + 1)2 = 42 = 16
Type yourWhich choice is the explicit formula for the following geometric sequence? answer here...
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.
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.
Finding the nth term is much simpler than it seems. For example, say you had the sequence: 1,4,7,10,13,16 Sequence 1 First we find the difference between the numbers. 1 (3) 4 (3) 7 (3) 10 (3) 13 (3) 16 The difference is the same: 3. So the start of are formula will be 3n. If it was 3n, the sequence would be 3,6,9,12,15,18 Sequence 2 But this is not our sequence. Notice that each number on sequence 2 is 2 more than sequence 1. this means are final formula will be: 3n+1 Test it out, it works!