Asked by Stinkrat
Instructions
Suppose we have a sequence
(1), (2, 3), (4, 5, 6). (7, 8, 9, 10), (11, 12, 13, 14, 15), ...
where (1) is the first element of the sequence, (2, 3) is the second, etc. What is the first number in the 100th element of the sequence? Be sure to do more than answer the problem. Could you use this problem with your students? What grade level would be appropriate? Could you adapt the problem to make it appropriate for at least four different grade levels?
Suppose we have a sequence
(1), (2, 3), (4, 5, 6). (7, 8, 9, 10), (11, 12, 13, 14, 15), ...
where (1) is the first element of the sequence, (2, 3) is the second, etc. What is the first number in the 100th element of the sequence? Be sure to do more than answer the problem. Could you use this problem with your students? What grade level would be appropriate? Could you adapt the problem to make it appropriate for at least four different grade levels?
Answers
Answered by
Steve
the sum of the 1st n elements is
1,3,6,10,15,...
it will be a cubic, since it's the sum of n(n+1)/2 for n terms
Sum of 1st n triangular numbers is
n(n+1)(n+2)/6
sum of 1st 99 elements is thus (99)(100)(101)/6 = 166650
So, the 100th element begins with the number 166651
since this involves proof by induction, I'd say algebra 2 would be a good place to present it.
1,3,6,10,15,...
it will be a cubic, since it's the sum of n(n+1)/2 for n terms
Sum of 1st n triangular numbers is
n(n+1)(n+2)/6
sum of 1st 99 elements is thus (99)(100)(101)/6 = 166650
So, the 100th element begins with the number 166651
since this involves proof by induction, I'd say algebra 2 would be a good place to present it.
Answered by
Stinkrat
Thanks Steve
What do you mean by its a cubic?
What do you mean by its a cubic?
Answered by
Steve
hmm. since we're talking about polynomials here, I figured it'd be clear that I meant a 3rd-degree polynomial. As opposed to a quadratic or linear.
Answered by
angel
3000+40,0000
Answered by
Justin
720
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