Sum of the first 41 square numbers

We define square numbers as numbers that when squared will equal a whole number. Thus, the list of the first square numbers starts with 1, 4, 9, 16, and so on.

What is the sum of the first 41 square numbers, you ask? Here we will give you the formula to calculate the first 41 square numbers and then we will show you how to calculate the first 41 square numbers using the formula.

The formula to calculate the first n square numbers is displayed below:

		n(n + 1) × (2(n) + 1)
		6

To calculate the sum of the first 41 square numbers, we enter n = 41 into our formula to get this:

		41(41 + 1) × (2(41) + 1)
		6

First, calculate each section of the numerator: 41(41 + 1) equals 1722 and (2(41) + 1) equals 83. Therefore, the problem above becomes this:

		1722 × 83
		6

Next, we calculate 1722 times 83 which equals 142926. Now our problem looks like this:

		142926
		6

Finally, divide the numerator by the denominator to get our answer:

142926 ÷ 6 = 23821

There you go. The sum of the first 41 square numbers is 23821.

You may also be interested to know that if you list the first 41 square numbers 1, 2, 9, etc., the 41st square number is 1681.

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