Does the same method used to find sum of the coefficients for a binomial hold here?
2026-03-25 17:44:45.1774460685
What is the sum of the coefficients in the expansion of $(x+y+w+z)^{20}$
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That will be $(1+1+1+1)^{20} = 4^{20}$.
You just set $x=y=z=w=1$ in the formula, and simplify.