Achild 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.

  f(N) = ((N+1)^2 -((N+1) mod 2))/4

Step-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

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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

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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 otherwise

another might be ...

  even(N) = (1 +(-1)^N)/2

So, 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

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A graph of this function for 1 to 10 blocks on the first row is attached.

answer

step-by-step explanation: