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December 18, 2014

December 18, 2014

Posted by **Stinkrat** on Saturday, February 9, 2013 at 4:04pm.

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?

- Math -
**Steve**, Saturday, February 9, 2013 at 4:15pmthe 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.

- Math -
**Stinkrat**, Saturday, February 9, 2013 at 5:37pmThanks Steve

What do you mean by its a cubic?

- Math -
**Steve**, Saturday, February 9, 2013 at 11:07pmhmm. 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.

- Math -
**angel**, Sunday, February 10, 2013 at 4:28pm3000+40,0000

- Math -
**Justin**, Monday, February 11, 2013 at 5:21pm720

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