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 used in the design.

Answer:   f(N) = ((N+1)^2 -((N+1) mod 2))/4Step-by-step explanation:The number of blocks is the sum of an arithmetic sequence with a common difference of -2. That sum is the product of the average number of blocks in a row and the number of rows.For N odd:The number in the base row is N, and the number in the top row is 1. The average number of blocks in a row is (N+1)/2. The number of rows is (N+1)/2, so the total number of blocks in the design is ...   f(N) = (N+1)^2/4__For N even:The number in the base row is N, and the number in the top row is 2. The average number of blocks in a row is (N+2)/2. The number of rows is N/2, so the total number of blocks in the design is ...   f(N) = ((N+1)^2 -1)/4__Combined formula:The difference between these formulas is an additive value of -1 when N is even. There are some different ways that an even/odd function can be written. One is ...   even(N) = (N+1) mod 2 . . . . . 1 when N is even; 0 otherwiseanother might be ...   even(N) = (1 +(-1)^N)/2So, for any integer N, the formula for the number of blocks in the design could be ...   f(N) = ((N+1)^2 -((N+1) mod 2))/4_____A graph of this function for 1 to 10 blocks on the first row is attached.