If \(f(x) = x^{n-1} + x^{n-2} + x^{n-3} + \ldots + 1\), then what is \(f'(2)\) equal to?
\(n2^{n-1} - 2^n + 1\)
\(f(x)\) is the geometric sum \(x^{n-1}+x^{n-2}+\ldots+1\), so \(f'(x) = (n-1)x^{n-2}+(n-2)x^{n-3}+\ldots+1\). Using the standard identity \(\sum_{i=1}^{n-1} i\,x^{i-1} = \dfrac{1-n x^{n-1}+(n-1)x^n}{(1-x)^2}\) and substituting \(x=2\), this simplifies to \(f'(2) = n\cdot 2^{n-1} - 2^n + 1\) (verified for \(n=2\), \(f'(2)=1\), and \(n=3\), \(f'(2)=5\)).
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