Question 12
If $$1 - x + x^2 - x^3 + \cdots + x^{20}$$ is rewritten in the form $$a_0 + a_1(x-4) + a_2(x-4)^2 + \cdots + a_{20}(x-4)^{20}$$, where $$a_0, a_1, \ldots, a_{20}$$ are all real numbers, the value of $$a_0 + a_1 + a_2 + \cdots + a_{20}$$ is
Solution
Substituting $$x - 4 = 1$$, that is $$x = 5$$, makes the right side equal to $$a_0 + a_1 + \cdots + a_{20}$$. The left side is a geometric series equal to $$\frac{1+x^{21}}{1+x}$$. At $$x = 5$$ this gives $$\frac{1+5^{21}}{6}$$.
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