Let $n \in \mathbb{N}^*$ and
$$T _ { n } ( X ) = \sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } ( - 1 ) ^ { p } \binom { n } { 2 p } X ^ { n - 2 p } \left( 1 - X ^ { 2 } \right) ^ { p }.$$
By expanding $( 1 + x ) ^ { n }$ for two appropriately chosen real numbers $x$, show that
$$\sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } \binom { n } { 2 p } = 2 ^ { n - 1 }.$$