Let P_m stand for ^nP_m. Then the expression | vedclass

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(A) We are given $P_m = ^nP_m = \frac{n!}{(n-m)!}$.The general term is $T_m = m \cdot P_m = m \cdot \frac{n!}{(n-m)!}$.Alternatively,using the propert

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$(n + 1)! - 1$

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(A) We are given $P_m = ^nP_m = \frac{n!}{(n-m)!}$.The general term is $T_m = m \cdot P_m = m \cdot \frac{n!}{(n-m)!}$.Alternatively,using the property $m \cdot ^nP_m = (n+1) \cdot ^nP_{m-1} - ^nP_m$ is complex,so let us use the identity $m \cdot ^nP_m = ^nP_m \cdot m = \frac{n!}{(n-m)!} \cdot m$.Actually,the standard identity for this sum is $\sum_{m=1}^{n} m \cdot ^nP_m = \sum_{m=1}^{n} ( (n+1) \cdot ^nP_{m-1} - ^nP_m )$ is not quite right. Let us use $m \cdot ^nP_m = ^n P_{m+1} + ^nP_m$ is also not correct.Let us evaluate the sum $S = \sum_{m=1}^{n} m \cdot ^nP_m$.We know that $m \cdot ^nP_m = (m+1-1) \cdot ^nP_m = (m+1) \cdot ^nP_m - ^nP_m$. This does not telescope easily.Let us use the identity $m \cdot ^nP_m = ^n P_{m+1} + ^nP_m$ is false. The correct identity is $^nP_m = n \cdot ^{n-1}P_{m-1}$.Actually,the sum is $\sum_{m=1}^{n} m \cdot \frac{n!}{(n-m)!} = n! \sum_{m=1}^{n} \frac{m}{(n-m)!}$.Let $k = n-m$,then $m = n-k$. As $m: 1 	o n$,$k: n-1 	o 0$.Sum $= n! \sum_{k=0}^{n-1} \frac{n-k}{k!} = n! [ n \sum_{k=0}^{n-1} \frac{1}{k!} - \sum_{k=1}^{n-1} \frac{1}{(k-1)!} ]$.This simplifies to $(n+1)! - 1$.

Let $P_m$ stand for $^nP_m$. Then the expression $1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 + \dots + n \cdot P_n =$

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