Consider $N\_1 = \sum\_{k=1}^K kr^k$ and $N\_2 = \sum\_{k=0}^K kr^k$ (assuming $r \neq 1$). 

These two equations are equivalent since the first term in $N\_2$ will always be zero due to the multiplication by $k$.

However, solving now:

$N\_1 = \sum\_{k=1}^K kr^k$

$N\_1 = r\sum\_{k=1}^K kr^{k-1}$

$N\_1 = r\left(\sum\_{k=1}^K r^{k}\right)^{'}$ (derivate trick)

Using geometric progression standard form:

$N\_1 = r\left(r\frac{1-r^{K}}{1-r}\right)^{'}= r\left(\frac{r-r^{K+1}}{1-r}\right)^{'}$

Now solving for $N\_2$:

$N\_2 = \sum\_{k=0}^K kr^k$

$N\_2 = r\sum\_{k=0}^K kr^{k-1}$

$N\_2 = r\left(\sum\_{k=0}^K r^{k}\right)^{'}$ (derivate trick)

Using geometric progression standard form:

$N\_2 = r\left(\frac{1-r^{K+1}}{1-r}\right)^{'}$ 

Clearly, $N\_1 \neq N\_2$ based on this expansion. Where is the flaw in the logic?