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Question

(D) Let $S = \sum_{0 \le i < j \le n} i \binom{n}{j}$.We can rewrite the sum by changing the order of summation:$S = \sum_{j=1}^{n} \binom{n}{j} \sum_

Vedclass · Accepted Answer

$n(n - 1) 2^{n-3}$

Vedclass · Answer

(D) Let $S = \sum_{0 \le i We can rewrite the sum by changing the order of summation:$S = \sum_{j=1}^{n} \binom{n}{j} \sum_{i=0}^{j-1} i$.The inner sum is the sum of the first $(j-1)$ integers: $\sum_{i=0}^{j-1} i = \frac{(j-1)j}{2}$.Thus,$S = \sum_{j=1}^{n} \binom{n}{j} \frac{j(j-1)}{2} = \frac{1}{2} \sum_{j=2}^{n} \frac{n!}{j!(n-j)!} j(j-1)$.Using the identity $j(j-1) \binom{n}{j} = n(n-1) \binom{n-2}{j-2}$,we get:$S = \frac{1}{2} \sum_{j=2}^{n} n(n-1) \binom{n-2}{j-2} = \frac{n(n-1)}{2} \sum_{j=2}^{n} \binom{n-2}{j-2}$.Let $k = j-2$,then the sum becomes $\sum_{k=0}^{n-2} \binom{n-2}{k} = 2^{n-2}$.Therefore,$S = \frac{n(n-1)}{2} \cdot 2^{n-2} = n(n-1) 2^{n-3}$.

$\sum_{0 \le i < j \le n} i \binom{n}{j}$ is equal to

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