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I asked a question here and got answer committing :

$$(1+yz^n)(1+y)^{n-1} - (1+yz)^n=\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}- \frac{1-z^n}{n}\bigg) \tag{1}$$and$$\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}= \frac{(z-1)^2}{n(n+1)}\bigg( \sum_{k=0}^{n-1} (k+1)z^k \bigg) \tag{2}$$

My try:

  1. Using binomial theorem, $$(1+yz^n)(1+y)^{n-1} - (1+yz)^n=(1+yz^n)\sum_{k=0}^{n-1}\binom{n-1}{k}y^{n-1-k}-\sum_{k=0}^n\binom{n}{k}(yz)^{n-k}$$
  2. Using this,$$\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}=(1-z)\cdot\bigg(\frac{1+z+\dots+z^{n-1}}{n}-\frac{1+z+\dots+z^{n-1}+z^n}{n+1}\bigg)$$ I can't go ahead in both the cases. Please help me.
Community
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Silent
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  • Equation (1) is correct. I do not know how this could help you : did you notice that (1-z^k)/k is the integral of z^(k-1) from z to 1 ? – Claude Leibovici Nov 02 '13 at 04:35

1 Answers1

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$1.$ \begin{align*} (1+yz^n)(1+y)^{n-1} - (1+yz)^n &= (1 + yz^n) \sum_{k=0}^{n-1} \binom{n-1}{k}y^k - \sum_{k=0}^{n} \binom{n}{k}(zy)^k \\ &= \sum_{k=1}^{n-1} y^k \left(\binom{n-1}{k} + z^n \binom{n-1}{k-1} - \binom{n}{k} z^k \right )\\ &= \sum_{k=1}^{n-1} \binom{n-1}{k-1} y^k \left( \frac{n-k}{k } + z^n - \frac{n}{k} z^k \right ) \\ &= \sum_{k=1}^{n-1} \binom{n-1}{k-1} y^k \left( \frac{n}{k}(1-z^k) - (1-z^n) \right )\\ \end{align*}

$2.$ $\displaystyle \sum_{k=0}^{n-1} (k+1)z^k$ is Arithmetico-geometric sequence can be evaluated as $\displaystyle \frac{n z^{n+1}-n z^n-z^n+1}{(z-1)^2}$ just by differentiating both sides of $\displaystyle \sum z^k = \frac{1-z^{n+1}}{1-z}$

Also to proceed from your step, \begin{align*} \frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1} &= (1-z)\cdot\left( \frac{1+z+\dots+z^{n-1}}{n}-\frac{1+z+\dots+z^{n-1}+z^n}{n+1}\right ) \\ &= \frac{(1-z)}{n(n+1)} \cdot \left( 1 + z + \dots + z^{n-1} - n z^n \right )\\ &= \frac{(1-z)}{n(n+1)} \cdot \left( \sum_{k=0}^{n-1} (k+1)z^k - \sum_{k=0}^{n-1}k z^k - \left( \sum_{k=0}^{n-1} (k+1)z^{k+1} - \sum_{k=0}^{n-1} k z^{k} \right ) \right )\\ \end{align*}

S L
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  • Sir, many many thanks. – Silent Nov 02 '13 at 05:03
  • @Sush added there. – S L Nov 02 '13 at 05:19
  • Many thanks,Sir. How did you derive $$\sum_{k=0}^{n-2}\binom{n-1}{k}y^{k+1}z^n=\sum_{k=0}^{n-1}\binom{n-1}{k-1}y^k z^{n}$$ in second step of 1. ? Also, in the original equality, we start sum from $k=1$ and not from $k=0$. (If we start summing from $k=0$, we are forced to divide by $0$). – Silent Nov 02 '13 at 05:58
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    @Sush $$\sum_{k=0}^{n-1} \binom{n-1}{k} z^n y^{k+1} = \sum_{k=1}^{n-1} \binom{n-1}{k-1} z^n y^{k} + z^n y^n$$. The last term get cancelled. – S L Nov 02 '13 at 06:09
  • Sir, will you just answer my last query? How is $$nz^n=\sum_{k=0}^{n-1} (k+1)z^{k+1} - \sum_{k=0}^{n-1}k z^k $$ – Silent Nov 02 '13 at 06:20
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    @Sush both has $n$ terms, the first term of second sum does not matter since $kz^k = 0$ when $k=0$. so first sum has n-terms and second sum has $n-1$ terms that get's cancelled. the the last term of first sum will only remain. – S L Nov 02 '13 at 06:22
  • All my doubts are cleared!! This is the effect of having good company!! Hope I will have so calm and helpful instructor some day!! Many many thanks again. – Silent Nov 02 '13 at 06:26
  • @Sush lol ... you are welcome :) – S L Nov 02 '13 at 06:27