Alternating Sums of Squares

Evaluate 12 – 22 + 32 – 42 + 52 – 62 + … + 1992.


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We have 1 + (32 – 22) + (52 – 42) + (72 – 62) + … + (1992 – 1982)
= 1 + 5 + 9 + 13 + … + 397.

This is an arithmetic sequence with first term 1, difference 4, last term 397. There are 100 terms.

The sum is n/2 · (first + last) = 100/2 · (1 + 397) = (50) · (398) = 19,900.

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