The sum of coefficients of integral powers of x in the binomial expansion (1−2√x)50 is
12(350+1)
Let Tr+1 be the general term in the expension of (1−2√x)50
∴ Tr+1=50Cr(1)50−r(−2x12)r=50Cr2rxr2(−1)r
For the integral power of x, r should be even integer.
∴ Sum of coefficients =∑25r=050C2r(2)2r
=12[(1+2)50+(1−2)50]=12(350+1)