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To solve this, we need to determine the total number of blocks \( N \) used in the triangular design based on whether \( n \) (the number of blocks in the base row) is odd or even.

In this setup:
- The base row has \( n \) blocks.
- Each subsequent row has 2 fewer blocks than the row below it, forming a triangle until the last row has either 1 block (if \( n \) is odd) or 2 blocks (if \( n \) is even).

### Analysis
1. **If \( n \) is odd**:
   The sequence of blocks per row would be \( n, n-2, n-4, \dots, 1 \).
   This forms an arithmetic sequence where:
   - First term \( a = n \),
   - Common difference \( d = -2 \),
   - Last term \( l = 1 \).

   Using the formula for the sum of an arithmetic sequence, \( S = \frac{k}{2} (a + l) \), where \( k \) is the number of terms:
   - We find \( k \) by solving \( n - 2(k-1) = 1 \), leading to \( k = \frac{n+1}{2} \).
   - Substituting values: 
     \[
     N = \frac{\frac{n+1}{2}}{2} \cdot (n + 1) = \frac{(n+1)^2}{4}
     \]

2. **If \( n \) is even**:
   The sequence of blocks per row would be \( n, n-2, n-4, \dots, 2 \).
   This forms an arithmetic sequence where:
   - First term \( a = n \),
   - Common difference \( d = -2 \),
   - Last term \( l = 2 \).

   We find \( k \) by solving \( n - 2(k-1) = 2 \), leading to \( k = \frac{n}{2} \).
   - Substituting values: 
     \[
     N =

A child places n cubic building blocks in a row to form the base of a triangular design. Each successive row contains two fewer blocks than the preceding row. Find a formula for the number of blocks N used in the design, depending on whether n is odd or even.

Learn how to calculate the total number of blocks N in a triangular building block design, where each row has two fewer blocks than the previous one. This solution provides formulas for cases when n is odd or even, using arithmetic sequences. Ideal for Grades 8-10.

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