I tried it by letting expression $1+2w+3w^2+\dots+nw^{n-1}= x$ and then multiplying $w$ both sides . I subtracted equation 1 from 2 but it does not seems to help me because i have just started learning this topic and i am weak in properties . Can anyone help me out here ?
Asked
Active
Viewed 9,314 times
2
-
Wait, why is that sum supposed to be $\frac{-n}{1-w}$? – Thomas Andrews Oct 13 '15 at 15:36
-
My mistake , edited it . – Seth Rollins Oct 13 '15 at 15:38
-
@Seth: What happen if you integrate the left side expression? – Piquito Oct 13 '15 at 15:42
-
See https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence – lab bhattacharjee Oct 13 '15 at 15:44
-
Well,if I don't understand anything in such general formula questions I normally take an example for example take w as first 2nd th root of unity, and then as cube root of unity as this will help you make or visualise or see a pattern and then you can prove the general formula by ways given here – Freelancer Oct 13 '15 at 15:45
3 Answers
4
Recall that $$\frac{z^n-1}{z-1} = 1+z+\cdots+z^{n-1}, \quad z \ne 1$$ so upon multiplying both sides by $z$ and differentiating, we obtain $$\frac{d}{dz}\left[\frac{z(z^n-1)}{z-1}\right] = 1+2z+3z^2+\cdots+nz^{n-1}.$$ After computing the derivative, evaluate the LHS for a primitive root of unity $z = e^{2\pi i/n}$.
Alternatively, by Abel summation: let $$f(z) = 1+2z+3z^2+\cdots+nz^{n-1},$$ so that $$\begin{align*} zf(z) - f(z) &= (z+2z^2+3z^3+\cdots+nz^n) - (1+2z+3z^2+\cdots+nz^{n-1}) \\ &= nz^n - (1+z+z^2+\cdots+z^{n-1}) \\ &= nz^n - \frac{z^n-1}{z-1}. \end{align*}$$ It immediately follows that $$f(z) = \frac{nz^n(z-1) - (z^n-1)}{(z-1)^2},$$ and the same substitution as in the above method easily results in $$f(w) = \frac{n(w-1)}{(w-1)^2} = \frac{n}{w-1}.$$
heropup
- 135,869
-
-
@Seth: I would consider the first one over the horizontal line fairly short, just the derivative of a geometric sum. – mathreadler Oct 13 '15 at 16:07
1
$$\sum_{k=0}^n w^k = \frac{w^{n+1}-1}{w-1}\implies \sum_{k=0}^nkw^{k-1} = \frac{(n+1)(w-1)w^n - \left(w^{n+1} - 1\right)}{\left(w-1\right)^2}$$ We know that $w^n = 1$, and so we can reduce this to be: $$\frac{(n+1)(w-1) - \left(w - 1\right)}{\left(w-1\right)^2} = \frac n{w - 1}$$
jameselmore
- 5,207
1
Applying twice your method of multiplying by $w$ and subtracting, and using $w^n=1$
$$1+2w+3w^2+\dots+nw^{n-1}= x$$ $$w+2w^2+3w^3+\dots+nw^{n}= xw$$ $$1+w+w^2+\dots w^{n-1}= x(1-w)+n$$ $$w+w^2+w^3+\dots+ w^{n}= x(1-w)w+nw$$ $$1-1=x(1-w)w+nw$$ $$x=\frac{-n}{1-w}$$ unless $w=1$ (prohibiting the division), in which case the sum is $n(n+1)/2$; you can ignore the possibility that $w=0$.
Henry
- 157,058