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

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(C) Let the given expression be $S = S_1 + S_2$,where $S_1 = (2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots$ up to $51$ term

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$1 + (52)!$

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(C) Let the given expression be $S = S_1 + S_2$,where $S_1 = (2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots$ up to $51$ terms) and $S_2 = (1! - 2! + 3! - \dots$ up to $51$ terms).We know that ${}^{n}P_{n-1} = n!$.Thus,$S_1 = (2 \cdot 1! - 3 \cdot 2! + 4 \cdot 3! - \dots + 52 \cdot 51!)$.Since $(n+1) \cdot n! = (n+1)!$,we have $S_1 = (2! - 3! + 4! - \dots + 52!)$.Now,$S_2 = (1! - 2! + 3! - 4! + \dots + 51!)$.Adding $S_1$ and $S_2$:$S = (2! - 3! + 4! - \dots + 52!) + (1! - 2! + 3! - 4! + \dots + 51!)$.All intermediate terms cancel out:$S = 1! + 52! = 1 + 52!$.

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

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