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An alternating sum of squares

EasyBrainteasers~3m

Evaluate 1002−992+982−972+⋯+22−12100^2 - 99^2 + 98^2 - 97^2 + \dots + 2^2 - 1^2.

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  1. Pair terms: 1002−992100^2-99^2, 982−97298^2-97^2, … and factor each as (a−b)(a+b)(a-b)(a+b).
  2. Each pair is a+ba+b, so the total is 1+2+⋯+1001+2+\dots+100.

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5050

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Asked at: Jane Street, Optiver

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