sum_{0 le i

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

Similar Questions

There were two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves exceeded the number of games that the men played with the women by $66$. The total number of participants is:

The value of $r$ for which $^{20}C_r \cdot ^{20}C_0 + ^{20}C_{r-1} \cdot ^{20}C_1 + ^{20}C_{r-2} \cdot ^{20}C_2 + \dots + ^{20}C_0 \cdot ^{20}C_r$ is maximum is:

In how many ways can five examination papers be arranged so that physics and chemistry papers never come together?

If the letters of the word $KRISNA$ are arranged in all possible ways and these words are written out as in a dictionary,then the rank of the word $KRISNA$ is

$^nC_r + 2^nC_{r-1} + ^nC_{r-2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module