Using the binomial theorem, we have:
$(1 + kx)^6 = \binom{6}{0} (1)^6 (kx)^0 + \binom{6}{1} (1)^5 (kx)^1 + \binom{6}{2} (1)^4 (kx)^2 + \binom{6}{3} (1)^3 (kx)^3 + \cdots$
Now, we only need the first four terms, so we can ignore the rest:
$(1 + kx)^6 =
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