The value of (2 cdot {}^{1}P_{0} - 3 cdot {}^ | vedclass

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(C) Let $S = S_{1} + S_{2}$,where $S_{1} = \sum_{n=1}^{51} (-1)^{n-1} (n+1) \cdot {}^{n}P_{n-1}$ and $S_{2} = \sum_{n=1}^{51} (-1)^{n-1} n!$. Since ${

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$1 + (52)!$

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(C) Let $S = S_{1} + S_{2}$,where $S_{1} = \sum_{n=1}^{51} (-1)^{n-1} (n+1) \cdot {}^{n}P_{n-1}$ and $S_{2} = \sum_{n=1}^{51} (-1)^{n-1} n!$. Since ${}^{n}P_{n-1} = n!$,we have $S_{1} = \sum_{n=1}^{51} (-1)^{n-1} (n+1) n! = \sum_{n=1}^{51} (-1)^{n-1} (n+1)!$. Expanding $S_{1} = 2! - 3! + 4! - \dots + (-1)^{50} (52)! = 2! - 3! + 4! - \dots + 52!$. Expanding $S_{2} = 1! - 2! + 3! - 4! + \dots + (-1)^{50} (51)! = 1! - 2! + 3! - 4! + \dots + (51)!$. Adding $S_{1}$ and $S_{2}$,the terms cancel out: $S = (1! - 2! + 3! - 4! + \dots + (51)!) + (2! - 3! + 4! - \dots + 52!) = 1! + 52! = 1 + 52!$.

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots \text{ up to } 51^{\text{th}} \text{ term}) + (1! - 2! + 3! - \dots \text{ up to } 51^{\text{th}} \text{ term})$ is equal to

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