Sum of coefficients of terms of even powers of x in(1+x+x2+x3)5 is

Given expansion (1+x+x2+x3)5 is Sum of coefficients of terms of even powers of x in(1+x+x2+x3)5 is  Let consider f(x)=(1+x+x2+x3)5  substitute x=1 then we get⇒f(1)=(1+1+1+1)5= (4)5  Similarly let consider f(x)=(1+x+x2+x3)5  substitute x=−1 then we get⇒f(−1)=(1−1+1−1)5=0 Sum of coefficients of even powers ofx =f(1)+f(−1)2                                                                            =45+02=29=512