OK, would the following be correct?
$\displaystyle \sum_{j=0}^{k+1}(k+1)_{j}$=
$\displaystyle (k+1)[k! + \sum_{j=0}^{k}(k)_{j}]$
where $\displaystyle (k)_j = _kP_j$
Dear oldguynewstudent,
When k=1,
$\displaystyle \sum^{k+1}_{j=0}{(k+1)_{j}}=\sum^{k+1}_{j=0}{~^{k+ 1}P_{j}}=5$
$\displaystyle (k+1)\left[k! + \sum_{j=0}^{k}(k)_{j}\right]=6$
Therefore, $\displaystyle \sum^{k+1}_{j=0}{(k+1)_{j}}\neq(k+1)\left[k! + \sum_{j=0}^{k}(k)_{j}\right]$
Hope this will help you.