\[\large \sum _{ n=0 }^{ 1729 }{ \frac { { \left( -1 \right) }^{ n }{ 1729 \choose n }}{ 4n+2 } ={ 2 }^{ A }\frac { B! }{ C!! } } \]

Find \(A+B+C\).

Note: The double factorial is defined as

For an

**even**positive integer, \(n!!=n\times (n-2)\times \cdots\times 4\times 2.\)For an

**odd**positive integer, \(n!!=n\times (n-2)\times \cdots\times 3\times 1.\)If \(n=0\), \(0!!=1.\)

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