What is the sum of the first 12 terms of the series 1, 4, 9, 16 ...

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The sum of the first 12 terms of the series consisting of the terms 1, 4, 9, 16 ... has to be determined.

From the terms that are given it can be seen that 1 = 1^2, 4 = 2^2, 9 = 3^2 and 16 = 4^2. The nth term of the series is equal to n^2.

To determine the sum of the first 12 terms of this series use the formula for the sum of squares. The sum of the squares of the first n integers starting from 1 is given by `S_n = (n*(n+1)*(2n+1))/6`

This gives the sum of the first 12 terms of the given series as `(12*(12+1)*(2*12 + 1))/6` = 2*13*25 = 650

**The sum of the first 12 terms of the given series is 650**

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