There are infinitely many possible functions that can generate this sequence. One such is
Un = (n2 - 3n + 2)/2 = (n-2)*(n-1)/2
There are infinitely many possible functions that can generate this sequence. One such is
Un = (n2 - 3n + 2)/2 = (n-2)*(n-1)/2
There are infinitely many possible functions that can generate this sequence. One such is
Un = (n2 - 3n + 2)/2 = (n-2)*(n-1)/2
There are infinitely many possible functions that can generate this sequence. One such is
Un = (n2 - 3n + 2)/2 = (n-2)*(n-1)/2
There are infinitely many possible functions that can generate this sequence. One such is
Un = (n2 - 3n + 2)/2 = (n-2)*(n-1)/2
It is the sequence of first differences. If these are all the same (but not 0), then the original sequence is a linear arithmetic sequence. That is, a sequence whose nth term is of the form t(n) = an + b
7 - 4n where n denotes the nth term and n starting with 0
This is the Fibonacci sequence, where the number is the sum of the two preceding numbers. The nth term is the (n-1)th term added to (n-2)th term
The pattern for the sequence 0 0 1 3 6 is that each term is obtained by adding the previous term multiplied by its position in the sequence (starting from 1). In other words, the nth term is given by n*(n-1)/2.
The nth term of that series is (24 - 6n).
24 - 6n
This is an arithmetic progression. In general, If an A.P. has a first term 'a', and a common difference 'd' then the nth term is a + (n - 1)d. In the sequence shown in the question, the first term is 0 and the common difference is 5, therefore the nth term is, 0 + (n - 1)5. This can be rearranged to read : 5(n - 1) For example : the 7th term is 30 : 5(7 - 1) = 5 x 6 = 30.
It is the sequence of first differences. If these are all the same (but not 0), then the original sequence is a linear arithmetic sequence. That is, a sequence whose nth term is of the form t(n) = an + b
7 - 4n where n denotes the nth term and n starting with 0
18 - 6n
Ok, take the formula dn+(a-d) this is just when having a sequence with a common difference dn+(a-d) when d=common difference, a=the 1st term, n=the nth term - you have the sequence 2, 4, 6, 8... and you want to find the nth term therefore: dn+(a-d) 2n+(2-2) 2n Let's assume you want to find the 5th term (in this case, the following number in the sequence) 2(5) = 10 (so the fifth term is 10)
10-2x for x = 0, 1, 2, 3, ... Since the domain of an arithmetic sequence is the set of natural numbers, then the formula for the nth term of the given sequence with the first term 10 and the common difference -2 is an = a1 + (n -1)(-2) = 10 - 2n + 2 = 12 - 2n.
If the nth term is 8 -2n then the 1st four terms are 6, 4, 2, 0 and -32 is the 20th term number
This is the Fibonacci sequence, where the number is the sum of the two preceding numbers. The nth term is the (n-1)th term added to (n-2)th term
The pattern for the sequence 0 0 1 3 6 is that each term is obtained by adding the previous term multiplied by its position in the sequence (starting from 1). In other words, the nth term is given by n*(n-1)/2.
The nth term of that series is (24 - 6n).
t(n) = n(n - 3)