Using the identity r+11(rn)=n+11(r+1n+1),
we get Pn=r=0∑nr+1(−2)rCr=n+11s=1∑n+1(−2)s(sn+1).
Since (1−2)n+1=(−1)n+1=s=0∑n+1(−2)s(sn+1),
we have s=1∑n+1(−2)s(sn+1)=(−1)n+1−1.
For even n=2k: P2k=2(2k+1)1−(−1)=2k+11.
Thus P2n1=2n+1, and n=1∑25(2n+1)=2⋅225⋅26+25=675.