Sum of the coefficients of(1+x−3x2)171=
1
2
3
−1
Given expansion (1+x−3x2)171 is
Sum of the coefficients of (1+x−3x2)171
Let consider f(x)=(1+x−3x2)171 substitute x=1 then we get
⇒f(1)=(1+1−3)171
⇒f(1)=(−1)171=−1