Splitting sum of permutation

Oct 2009
255
20
St. Louis Area
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 \(\displaystyle (k)_j = _kP_j\)
 
Dec 2009
872
381
1111
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.
 
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