Let an=∑r=0n 1 nCr2 and bn=∑r=0n (n−r) nCr2
then bn equals
an
nan
n2an
0
We have
bn=∑r=0n n−r nCr2=∑r=0n n−r nCn−r2=∑r=0n r nCr2
Thus,
bn+bn=∑r=0n n−rnCr2+∑r=0n rnCr2=n∑r=0n 1 nCr2=nan
⇒ bn=n2an