A child places n cubic building blocks in a row to form the base of a triangular design (see figure). Each successive row contains two fewer blocks than the preceding row. Find a formula for the number of blocks N used in the design. (Hint: The number of building blocks in the design depends on whether n is odd or even.)
If n is odd, N =
If n is even, N =
n is even in the block
To find a formula for the number of blocks used in the triangular design, we need to consider two cases: when `n` is odd and when `n` is even.
1. When `n` is odd:
In this case, the number of blocks used in each row forms an arithmetic sequence starting from `n` and decreasing by 2 in each subsequent row. The last row will have 1 block. So, the number of blocks used in the design is the sum of the arithmetic sequence.
The formula for the sum of an arithmetic sequence is given by: Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term.
Since the first term is `n`, the last term is `1`, and the common difference is `-2`, the formula becomes:
N = (n/2)(n + 1).
2. When `n` is even:
In this case, the number of blocks used in each row forms an arithmetic sequence starting from `n` and decreasing by 2 in each subsequent row. The last row will have 2 blocks. So, the number of blocks used in the design is the sum of the arithmetic sequence, plus 1 (to account for the extra block at the end).
Using the same formula for the sum of an arithmetic sequence, the last term for this case becomes `2`, and the formula becomes:
N = (n/2)(n + 2) + 1.
So, if `n` is odd, the formula for the number of blocks used in the design is N = (n/2)(n + 1). And if `n` is even, the formula is N = (n/2)(n + 2) + 1.
If n is odd, the number of blocks N used in the design can be found using the formula:
N = 1 + 3 + 5 + ... + (2n - 1)
We can notice that this series represents the sum of the first n odd numbers.
If we denote the sum of the first n odd numbers as S(n), then we have:
S(n) = 1 + 3 + 5 + ... + (2n - 1)
To find the formula for S(n), we can observe that each term in the series can be expressed as 2i - 1, where i is the position of the term.
So, we have:
S(n) = (2*1 - 1) + (2*2 - 1) + (2*3 - 1) + ... + (2n - 1)
= 1 + 2 + 3 + ... + n
= n(n + 1)/2
Therefore, if n is odd, the number of blocks N used in the design is given by:
N = n(n + 1)/2.
If n is even, the number of blocks N used in the design can be found using the formula:
N = 2 + 4 + 6 + ... + 2n
Similar to before, we notice that this series represents the sum of the first n terms of the even numbers.
If we denote the sum of the first n even numbers as S(n), then we have:
S(n) = 2 + 4 + 6 + ... + 2n
Again, we can observe that each term in the series can be expressed as 2i, where i is the position of the term.
So, we have:
S(n) = 2*1 + 2*2 + 2*3 + ... + 2n
= 2(1 + 2 + 3 + ... + n)
= 2 * n(n + 1)/2
= n(n + 1)
Therefore, if n is even, the number of blocks N used in the design is given by:
N = n(n + 1).
Have you tried laying out a few rows? I'm sure you will see a pattern.
You will be adding up sequences of even or odd numbers...