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6.Permutation and Combination
normal
$\mathop \sum \limits_{0 \le i < j \le n} i\left( \begin{array}{l}
n\\
j
\end{array} \right)$ is equal to
A
$n^22^{n-1}$
B
$(n^2 -1)2^{n-1}$
C
$(n-1)^22^n$
D
$n(n-1)2^{n-3}$
Solution
$\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)} $
$=\frac{n(n-1)}{2} 2^{n-2}=n(n-1) 2^{n-3}$
Standard 11
Mathematics
