I would like to know what is the formula por this sequence please
-
The first answer is the right way of looking at it, but isn't an actual formula.
Notice that the first 1 occurs in the 2nd position, the next 1 occurs in the 5th position, then in the 9th position, the 14th position, and so on. The pattern is the position of the first 1 is 2, you add 3 to get to the position of the next 1, add 4 to get to the position of the following 1, and so on. So you add 1 more each time to get to the position of the subsequent 1.
Using that pattern, you can see that the sequence of positions with 1's (2,5,9,14,...) can be defined by a recurrence relation:
T(0) = 2, T(n) = T(n-1) + n + 2 for n > 0
Using that recurrence relation, write out the first few terms:
T(0) = 2
T(1) = T(0) + 1 + 2 = 2 + 1 + 2 = 5
T(2) = T(1) + 2 + 2 = 5 + 2 + 2
T(3) = T(2) + 3 + 2 = 5 + 2 + 2 + 3 + 2
T(4) = T(3) + 4 + 2 = 5 + 2 + 2 + 3 + 2 + 4 + 2
T(5) = T(4) + 5 + 2 = 5 + 2 + 2 + 3 + 2 + 4 + 2 + 5 + 2
By now the pattern should be clear: Starting at n=3, you add 5 and n 2's and each number from 3 to n. The sum of the numbers from 1 to n is given by the formula n(n+1)/2, so the sum of the numbers from 3 to n is n(n+1)/2 - 1 - 2, which is
n(n+1)/2 - 3. Adding 5 and n 2's to that, you get:
T(n) = n(n+1)/2 - 3 + 5 + 2n
T(n) = (1/2)n² + (1/2)n + 2 + 2n
T(n) = (1/2)n² + (5/2)n + 2
T(n) = (1/2)(n² + 5 + 4)
T(n) = (1/2)(n+1)(n+4)
Now, that was based on the pattern starting at n=3, so we need to verify that the formula holds for n=0, n=1 and n=2:
T(0) = (1/2)(0+1)(0+4) = 2
T(1) = (1/2)(1+1)(1+4) = 5
T(2) = (1/2)(2+1)(2+4) = 9
Those are indeed the correct numbers, so our formula holds for all n.
Thus, the kth term of the original sequence is 1 if k is of the form (1/2)(n+1)(n+4) for some non-negative integer n, and 0 otherwise.
Hope that helps :)
Notice that the first 1 occurs in the 2nd position, the next 1 occurs in the 5th position, then in the 9th position, the 14th position, and so on. The pattern is the position of the first 1 is 2, you add 3 to get to the position of the next 1, add 4 to get to the position of the following 1, and so on. So you add 1 more each time to get to the position of the subsequent 1.
Using that pattern, you can see that the sequence of positions with 1's (2,5,9,14,...) can be defined by a recurrence relation:
T(0) = 2, T(n) = T(n-1) + n + 2 for n > 0
Using that recurrence relation, write out the first few terms:
T(0) = 2
T(1) = T(0) + 1 + 2 = 2 + 1 + 2 = 5
T(2) = T(1) + 2 + 2 = 5 + 2 + 2
T(3) = T(2) + 3 + 2 = 5 + 2 + 2 + 3 + 2
T(4) = T(3) + 4 + 2 = 5 + 2 + 2 + 3 + 2 + 4 + 2
T(5) = T(4) + 5 + 2 = 5 + 2 + 2 + 3 + 2 + 4 + 2 + 5 + 2
By now the pattern should be clear: Starting at n=3, you add 5 and n 2's and each number from 3 to n. The sum of the numbers from 1 to n is given by the formula n(n+1)/2, so the sum of the numbers from 3 to n is n(n+1)/2 - 1 - 2, which is
n(n+1)/2 - 3. Adding 5 and n 2's to that, you get:
T(n) = n(n+1)/2 - 3 + 5 + 2n
T(n) = (1/2)n² + (1/2)n + 2 + 2n
T(n) = (1/2)n² + (5/2)n + 2
T(n) = (1/2)(n² + 5 + 4)
T(n) = (1/2)(n+1)(n+4)
Now, that was based on the pattern starting at n=3, so we need to verify that the formula holds for n=0, n=1 and n=2:
T(0) = (1/2)(0+1)(0+4) = 2
T(1) = (1/2)(1+1)(1+4) = 5
T(2) = (1/2)(2+1)(2+4) = 9
Those are indeed the correct numbers, so our formula holds for all n.
Thus, the kth term of the original sequence is 1 if k is of the form (1/2)(n+1)(n+4) for some non-negative integer n, and 0 otherwise.
Hope that helps :)
-
Each subsequent group of zeros has 1 more zero than the previous group.