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I am really stuck on how to do this. It is a step I need to do for an inductive proof. I have $$\frac{1}{n(n-1)} $$

Do I set it up like this: $\frac{ A}{n} + \frac{B}{n-1}$ ? $$1= A(n-1) + B(n)$$ How do I solve for $A$ and $B$

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You have the fraction $$\frac{1}{n(n-1)}$$ And you want it to be $$\frac{1}{n(n-1)} =\frac{A}{n}+\frac{B}{n-1}$$ Now you expand it and get $$\frac{A(n-1)+Bn}{n(n-1)}=\frac{1}{n(n-1)}$$ Now you compare the coefficients, which give you that $$A(n-1) + Bn=0\cdot n + 1 $$ 2 polynomials are the same iff all their coefficients are the same. $$(A+B) \cdot n =0 \cdot n$$ and $$ -A = 1$$ This gives $B=1$ and $A=-1$ hence $$\frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}$$

Instead of the coefficient comparism you can $k+1$ values for the polynomial where $k$ is the degree of the polynomial. Showing that this works to is a result of polynom interpolation, that works at least for $\mathbb{R}$.

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  • $\begingroup$ So do I replace -1 for A to solve for B? $\endgroup$ – Marla Apr 3 '13 at 17:27
  • $\begingroup$ Do the same as @Dominic suggested till you get $$A(n-1)+Bn=1$$ Now set $n=1$ above so $A(1-1)+B=1$ so $B=1$. Set $n=0$ above so $A(0-1)+0=1$ so $A=-1$. $\endgroup$ – mrs Apr 3 '13 at 18:39

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