Does anyone know how it is possible to lose the n element in the index of the summation, while remaining equal ?
$$ \sum_{k=0}^{n} \binom{n}{k}q^kp^{n-k}(n-k)^1 = np\sum_{k=0}^{n-1} \binom{n-1}{k}q^kp^{n-k-1} $$
I understand the following steps
$$ \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}q^kp^{n-k+1-1}(n-k)^1 $$
$$ =\sum_{k=0}^{n} \frac{n!}{k!(n-k)^{1}(n-k-1)!}q^kp^{n-k-1}p^{1}(n-k)^1 $$