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?
3000+40,0000
720
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.
Thanks Steve
What do you mean by its a cubic?
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.
To find the first number in the 100th element of the given sequence, we need to identify the pattern in which the numbers are arranged.
Looking closely at the sequence, we can observe that each element consists of consecutive integers starting from a specific number. The first element is (1), the second element is (2, 3), the third element is (4, 5, 6), and so on.
Let's analyze further. In each element, the number of integers increases by one compared to the previous element. For example, the second element has two integers (2, 3), while the first element has only one integer, which is 1.
Another pattern we can notice is that the starting number of each element increases by the number of integers present in that element. For instance, the second element starts with the number 2, which is one more than the last number in the first element. Similarly, the third element starts with 4, which is one more than the last number in the second element.
Based on these patterns, we can conclude that the first number in the n-th element can be calculated using the following formula:
First number = 1 + 2 + 3 + ... + (n-2) + (n-1)
This formula adds up all the numbers from 1 to (n-1) to give us the first number in the n-th element.
Now, let's apply this formula to find the first number in the 100th element:
First number = 1 + 2 + 3 + ... + 98 + 99
To calculate this sum, we can use the formula for the sum of consecutive integers:
Sum = (n * (n + 1)) / 2
Using this formula, we can find the sum:
Sum = (99 * (99 + 1)) / 2 = (99 * 100) / 2 = 4950
Therefore, the first number in the 100th element of the sequence is 4950.
Now, let's consider how we can use this problem with students of different grade levels:
1. Elementary school (Grade 3-5): Students at this level can practice identifying number patterns and extending the sequence to find the first number in any given element. The problem can be used as an introduction to sequences and patterns.
2. Middle school (Grade 6-8): Students at this level can explore the concept of arithmetic sequences. They can derive the formula for the sum of consecutive integers and apply it to find the first number in any element. This problem can be used to reinforce their understanding of arithmetic sequences.
3. High school (Grade 9-12): Students at this level can apply advanced techniques like summation notation to find the first number in any element. They can also explore the concept of series and discuss convergence. This problem can serve as a challenging exercise to test their skills in algebra and series.
4. College/University: This problem can be used as a starting point to discuss the concept of series and summation in higher-level mathematics courses, such as calculus or discrete mathematics. Students can analyze the sequence and derive general formulas to find the first number in any element.
By adapting the level of complexity in the explanation and the mathematical techniques used, we can make this problem suitable for various grade levels.