How to get $\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}- \frac{1-z^n}{n}\bigg)$"?

Question

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.

Answer 1

$1.$ \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*}