If n is a positive integer such that n ge 3,t | vedclass

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(A) The given series is $S = \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} (n-r)$.This can be written as $S = n \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} - \sum

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(A) The given series is $S = \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} (n-r)$.This can be written as $S = n \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} - \sum_{r=0}^{n-1} (-1)^r r \binom{n-1}{r}$.For the first part,$\sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} = (1-1)^{n-1} = 0$ for $n-1 \ge 1$ (i.e.,$n \ge 2$).For the second part,using the identity $r \binom{n-1}{r} = (n-1) \binom{n-2}{r-1}$,we have $\sum_{r=1}^{n-1} (-1)^r (n-1) \binom{n-2}{r-1} = (n-1) \sum_{k=0}^{n-2} (-1)^{k+1} \binom{n-2}{k} = -(n-1) (1-1)^{n-2}$.For $n > 2$,$(1-1)^{n-2} = 0$,so the sum is $0$.For $n=3$,the series is $1(3) - \frac{2}{1}(2) + \frac{2 \cdot 1}{2}(1) = 3 - 4 + 1 = 0$.

If $n$ is a positive integer such that $n \ge 3$,then the value of the sum to $n$ terms of the series $1 \cdot n - \frac{(n - 1)}{1!} (n - 1) + \frac{(n - 1)(n - 2)}{2!} (n - 2) - \frac{(n - 1)(n - 2)(n - 3)}{3!} (n - 3) + \dots$ is:

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