MathDB
$P(x)$ has real roots

Source: VNTST 2019 P2

April 7, 2019
algebrapolynomial

Problem Statement

For each positive integer nn, show that the polynomial: Pn(x)=k=0n2k(2n2k)xk(x1)nkP_n(x)=\sum _{k=0}^n2^k\binom{2n}{2k}x^k(x-1)^{n-k} has nn real roots.