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Let $f(x) = x^{2n} - x^{2n-1} + \cdots + x^4 - x^3 + x^2 - ax + b$

We wish to compute $f'(1)$. The way I did this was to differentiate term by term, and then pair the terms of the form $-(2m + 1)x^{2m} + (2m) x^{2m-1}$ to get $-1$ summed $n-1$ times to get a $1-n$ term. Considering the first and last terms, we see that $2n$ and $-a$ contribute a term. So we get $f'(1) = n+1-a$

My book just states that $f'(1) = n-1+2-a$. They might have a quicker way than my reasoning, and I am not sure where the $+2$ comes from. Does anyone see how the book did it?

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    $f(x)=-ax+b+\sum_{k=2}^{2n}(-1)^kx^k$ hence $f'(x)=-a+\sum_{k=2}^{2n}(-1)^kkx^{k-1}$ and $$f'(1)=-a+\sum_{k=2}^{2n}(-1)^kk=-a+2+\sum_{k=3}^{2n}(-1)^kk=-a+2-\sum_{j=2}^n(2j-1)+\sum_{j=2}^n(2j)$$ and we get the result.2011-10-06
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    Or maybe $f(x)=(-ax +b)-(1-x)+[1-x+x^2-x^3+\cdots +x^{2n}]$. The sum in square brackets is $\frac{1+x^{2n+1}}{1+x}$, easy to differentiate.2011-10-06

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