#math trivia #110 solution

Let n be the index of the term, starting at 0.  The terms follow the pattern:

  110 = (10-n)*(11+n) + n*(n+1)

In other words, for each n, we can say:

  110 is n*(n+1) more than (10-n)*(11+n) .

The formula can be checked by expanding it:

  (10-n)*(11+n) + n*(n+1) = 110+n-n^2 + n^2+n = 110

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