mathop sum limits_{0 le i

Question

(D) Let $S = \sum_{0 \le i < j \le n} i \binom{n}{j}$.We can rewrite the sum as $\sum_{j=1}^n \binom{n}{j} \sum_{i=0}^{j-1} i$.The inner sum is $\sum_

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $S = \sum_{0 \le i We can rewrite the sum as $\sum_{j=1}^n \binom{n}{j} \sum_{i=0}^{j-1} i$.The inner sum is $\sum_{i=0}^{j-1} i = \frac{(j-1)j}{2}$.Thus,$S = \sum_{j=1}^n \binom{n}{j} \frac{j(j-1)}{2} = \frac{1}{2} \sum_{j=2}^n j(j-1) \binom{n}{j}$.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 $S = \frac{n(n-1)}{2} \sum_{k=0}^{n-2} \binom{n-2}{k}$.Since $\sum_{k=0}^{m} \binom{m}{k} = 2^m$,we have $S = \frac{n(n-1)}{2} 2^{n-2} = n(n-1) 2^{n-3}$.

List-$I$	List-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$	$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$	$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$	$(III)$ $\frac{1}{2}(3^n + 1)$
	$(IV)$ $\frac{1}{2}(3^n - 1)$

$\mathop \sum \limits_{0 \le i < j \le n} i \binom{n}{j}$ is equal to

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