Question

The absolute value of the sum of all the coefficients in the binomial expansion of ${(x^2+x-3)}^{319}$

Answer 1

Let
$(x^2+x-3{)}^{319}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{638}{x}^{638}$
Putting $x=1$ in above equation, we get
$(1+1-3{)}^{319}=a_0+a_1+a_2+\cdots +a_{638}\Rightarrow a_0+a_1+a_2+\cdots +a_{638}=-1$
Hence required absolute value will be $1$