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 =