if Cr stand for nCr, the sum of the given series  2n2!n2!n!⋅C02−2C12+3C22−…+(−1)n(n+1)Cn2,where n is an even positive integer, is

We have,C02−2C12+3C22−…+(−1)n(n+1)Cn2=C02−C12+C22−…+(−1)nCn2−C12−2C22+3C32−…+(−1)nnCn2=(−1)n/2n!n2!n2!⋅−(−1)n/2−1⋅12nn!n2!n2!=(−1)n2⋅n!n2!n2!⋅1+n2Therefore, the value of the given expression is 2n2!n2!n!×(−1)n2⋅(n)!n2!n2!1+n2=(−1)n2(2+n)