mathop sum limits_{0 le i

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Question

$\sum\limits_{0 \le i < j \le n} i \left( {\frac{n}{j}} \right) = \frac{{n\left( {n - 1} \right)}}{2}\sum\limits_{k = 0}^n {\left( {\frac{{n - 2}}{

Vedclass · Accepted Answer

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

Vedclass · Answer

$\sum\limits_{0 \le i < j \le n} i \left( {\frac{n}{j}} \right) = \frac{{n\left( {n - 1} \right)}}{2}\sum\limits_{k = 0}^n {\left( {\frac{{n - 2}}{k}} \right)} $

$=\frac{n(n-1)}{2} 2^{n-2}=n(n-1) 2^{n-3}$

$\mathop \sum \limits_{0 \le i < j \le n} i\left( \begin{array}{l}
n\\
j
\end{array} \right)$ is equal to

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$\mathop \sum \limits_{0 \le i < j \le n} i\left( \begin{array}{l} n\\ j \end{array} \right)$ is equal to