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I tried to solve it in the simple way by adding on both sides $(n+1)w^n$ but I was unsuccessful

user170231
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Alexandre Barros
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  • Can you elaborate on this "simple way"? How does it differ from the usual steps in an induction proof? – user170231 Dec 19 '23 at 16:22
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    Doesn't work for $n=w=1$. – peterwhy Dec 19 '23 at 16:33
  • Be very careful in the induction step as $w$ itself depends on $n$, so there is some bad notation involved.

    Instead you should consider $w_n=\mathop{\text{cis}}(\frac{2\pi}{n})$ to avoid such confusion, but then the main problem is that $w_{n+1}$ and $w_n$ are related only by a very inconvenient relation, which is $w_{n+1}^{n+1}=w_{n}^{n}$

     – b00n heT Dec 19 '23 at 16:43
  • Please do not answer questions in the comments. – Xander Henderson Dec 19 '23 at 17:08

1 Answers1

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As $w=e^{\frac{2\pi i}n}$, so $w^{n}=1$. Now by the formula of the G.P series, we can write $ \frac {x^{n+1}-1}{x-1} = 1+x+x^{2}+ \cdots +x^{n−1}$, by derivative both sides we get $\frac{(x-1)(n+1)x^{n} - (x^{n+1}-1)}{(x-1)^{2}} = 1+2x+3x^2+⋯+nx^{n−1}$(Here this derivative is possible because the above function is defined on $\mathbb{R} \setminus \{1\}$). If you put $x=w$ on both sides, you will get the answer as $w \neq 1$ and $w^{n}=1$.

Afntu
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  • Can you prove this result by induction? – Alexandre Barros Dec 21 '23 at 04:24
  • I don't think so because here, If you want to show this by induction on n, then if you assume this result is true for n and then want to prove for n+1, your w is not the same with the induction step because then w is the (n+1)the root of unity, but at first, it was the nth root of unity. GP series, which I used, requires an induction proof, as this result holds by induction, then derivative and putting the value w on it. – Afntu Dec 21 '23 at 13:18