If you want to ask questions about the "following", then I suggest that you make sure that there is something that is following.
Formula for the nth term of general geometric sequence tn = t1 x r(n - 1) For n = 2, we have: t2 = t1 x r(2 - 1) t2 = t1r substitute 11.304 for t2, and 2.512 for t1 into the formula; 11.304 = 2.512r r = 4.5 Check:
The formula for height depends on the context. There is no simple formula for the height of a person. Formulae for the height of a geometric shape depends on what information about the shape is given.
The given sequence is an arithmetic sequence with a common difference of -4. To find the nth term formula, we first determine the first term, which is 100. The nth term formula for an arithmetic sequence is given by: a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, n is the term number, and d is the common difference. Therefore, the nth term formula for this sequence is a_n = 100 - 4(n-1) or a_n = 104 - 4n.
The simplest formula isUn = (-8611*n^2 + 34477*n - 25082)/2 for n = 1, 2, 3.
what is the recursive formula for this geometric sequence?
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.
un = u0*rn for n = 1,2,3, ... where r is the constant multiple.
-7
It look like a Fibonacci sequence seeded by t1 = 2 and t2 = 1. After that the recursive formula is simply tn+1 = tn-1 + tn.
If you want to ask questions about the "following", then I suggest that you make sure that there is something that is following.
In this case, 22 would have the value of 11.
The formula to find the sum of a geometric sequence is adding a + ar + ar2 + ar3 + ar4. The sum, to n terms, is given byS(n) = a*(1 - r^n)/(1 - r) or, equivalently, a*(r^n - 1)/(r - 1)
In order to answer the question is is necessary to know what the explicit formula was. But, since you have not bothered to provide that information, the answer is .
Yes, it can both arithmetic and geometric.The formula for an arithmetic sequence is: a(n)=a(1)+d(n-1)The formula for a geometric sequence is: a(n)=a(1)*r^(n-1)Now, when d is zero and r is one, a sequence is both geometric and arithmetic. This is because it becomes a(n)=a(1)1 =a(1). Note that a(n) is often written anIt can easily observed that this makes the sequence a constant.Example:a(1)=a(2)=(i) for i= 3,4,5...if a(1)=3 then for a geometric sequence a(n)=3+0(n-1)=3,3,3,3,3,3,3and the geometric sequence a(n)=3r0 =3 also so the sequence is 3,3,3,3...In fact, we could do this for any constant sequence such as 1,1,1,1,1,1,1...or e,e,e,e,e,e,e,e...In general, let k be a constant, the sequence an =a1 (r)1 (n-1)(0) with a1 =kis the constant sequence k, k, k,... and is both geometric and arithmetic.
It is often possible to find an explicit formula that gives the same answer as a given recursive formula - and vice versa. I don't think you can always find an explicit formula that gives the same answer.
You need to know two numbers to completely describe the geometric sequence: the starting number, and the ratio between each number and the previous one. When you use recursion, you always need a "base case", otherwise, the recursion will repeat without end. In words, if "n" is 1, the result is the starting term. Otherwise, it is the ratio times the "n-1"th term. The following version is appropriate for a programming language (written here in pseudocode, i.e., not for a specific language): function geometric(starting_number, ratio, term) if term = 1: result = starting_number else: result = ratio * geometric(starting_number, ratio, term - 1)