An efficient solution is based on the below recursive formula for even Fibonacci Numbers. The third triangle has another row with 3 extra dots making 1 2 3 6. In the general case the sum of the first k odd numbers is the sum of the first k even numbers minus k.
That is k k1 - k k2.
2 3 5 7 11 where n is a natural number. It is simply the number of dots in each triangular pattern. 99 100 5050 step 3 Divide the sum by 100 Average 5050100 505 Thus 505 is an average of even numbers between 1 and 100. In the general case the sum of the first k odd numbers is the sum of the first k even numbers minus k.
