Let $$f(n)=\sum_{r=0}^{\big\lfloor{\frac{n}{2}\big\rfloor}} {{n-r} \choose r}$$
Prove,
$$f(x-2)+f(x-1)=f(x)$$
I don't know where to start. An someone help me solving this?
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Is $x=n$? In any case, induction seems natural. – lulu Feb 10 '23 at 17:47
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Induction is unnecessary; write down the sums and try to simplify. If you don't know where to start, look at some small cases and see what happens (don't simplify the whole thing to a number, that won't be educational; look at what the terms of the sums do and how they can be combined). – Misha Lavrov Feb 10 '23 at 17:48
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@MishaLavrov for example $x=20$, $f(x)$ will have $11$ terms and $f(x-2),f(x-1)$ will have 10 terms each so terms willn't match. And also the first and last terms of each can be simplified in terms of $f(x)$ but not the same for middle ones. – Krish Feb 10 '23 at 18:07
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@lulu can you prove by induction? – Krish Feb 10 '23 at 18:08
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You can, but I agree with @MishaLavrov that it isn't necessary. In any case, working it out carefully for small $n$ is the right place to start. – lulu Feb 10 '23 at 18:08
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@Krish So look at those terms! How can $$f(18)=1+17+120+455+1001+1287+924+330+45+1$$ and $$f(19)=1+18+136+560+1365+2002+1716+792+165+10$$ be manipulated to get $$f(20)=1+19+153+680+1820+3003+3003+1716+495+55+1$$ without having to simplify all three sums any further? – Misha Lavrov Feb 10 '23 at 18:18