For any positive integer n, the sum of the first n positive integers…
Although a formula is provided in this problem, we can easily solve it using a different formula:
sum = (average)(quantity)
Let’s first determine the average.
In any set of numbers in an arithmetic sequence, we can determine the average using the formula:
(1st number in set + last number in set)/2
Remember, we must average the first even integer in the set and the last even integer in the set. So we have:
(100 + 300)/2 = 400/2 = 200
Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive.
Two key points to recognize:
1) Because we are determining the number of “even integers” in the set, we must divide by 2 after subtracting our quantities.
2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must “add 1” after doing the subtraction.
quantity = (300 – 100)/2 + 1
quantity = 200/2 + 1
quantity = 101
Finally, we can determine the sum.
sum = 200 x 101
sum = 20,200
Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don’t want the total of all of the first 300 numbers. We’d have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we’d also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn’t very easy to use.