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Question Number 98844 by I want to learn more last updated on 16/Jun/20

$Please explain : \sum_{1 ⩽ i < j ⩽ n} ij = \underset{j = 2}{\sum^{n}} \frac{j (j - 1) j}{2}$ $I want to know how L . H . S = R . H . S$
	Answered by mr W last updated on 16/Jun/20

	$\sum_{1 ⩽ i < j ⩽ n} i j$ $= \underset{i = 1}{\sum^{n - 1}} (i \underset{j = i + 1}{\sum^{n}} j)$ $= \underset{i = 1}{\sum^{n - 1}} [i \times \frac{(n + i + 1) (n - i)}{2}]$ $o r$ $= \underset{i = 2}{\sum^{n}} (j \underset{i = 1}{\sum^{j - 1}} i)$ $= \underset{i = 2}{\sum^{n}} (j \times \frac{j \times (j - 1)}{2})$ $= \underset{j = 2}{\sum^{n}} \frac{j^{2} (j - 1)}{2}$
	Commented by I want to learn more last updated on 16/Jun/20

	$Thanks sir$
	Commented by I want to learn more last updated on 16/Jun/20

	Commented by I want to learn more last updated on 16/Jun/20

	$Sir my problems the red ink . How can we know it .$ $why we have : Σ j \underset{i = 1}{\sum^{j - 1}} i = \underset{j = 2}{\sum^{n}} . . . = \underset{j = 1}{\sum^{n}} ? ? ? ? ? ?$ $And what is the difference in$ $1 ⩽ i < j ⩽ n and 1 ⩽ i ⩽ j ⩽ n$
	Commented by mr W last updated on 16/Jun/20

	$1 ⩽ i < j ⩽ n m e a n s :$ $1 ⩽ i ⩽ n - 1 a n d i + 1 ⩽ j ⩽ n$ $o r$ $2 ⩽ j ⩽ n a n d 1 ⩽ i ⩽ j - 1$
	Commented by I want to learn more last updated on 16/Jun/20

	$Ohh . Thanks sir . I really appreciate$
	Commented by mr W last updated on 16/Jun/20

	Commented by mr W last updated on 16/Jun/20

	$r e d m a r k e d c o m b i n a t i o n s a r e$ $1 ⩽ i < j ⩽ n$ $t h e y c a n b e d e s c r i b e d a s$ $1 ⩽ i ⩽ n - 1 a n d i + 1 ⩽ j ⩽ n$ $o r a s$ $2 ⩽ j ⩽ n a n d 1 ⩽ i ⩽ j - 1$
	Commented by I want to learn more last updated on 16/Jun/20

	$wow, thanks sir for more details .$
	Answered by mathmax by abdo last updated on 16/Jun/20
	$Σ_(1≤i<j≤n) ij = Σ_(j=1) ^n (Σ_(i=1) ^(j−1) ij) =Σ_(j=1) ^n j((((j−1)j)/2)) =Σ_(j=1) ^n ((j^2 (j−1))/2) =(1/2)(Σ_(j=1) ^n j^3 −Σ_(j=1) ^n j^2 ) =(1/2){ ((n^2 (n+1)^2 )/4)−((n(n+1)(2n+1))/6)} =((n(n+1))/2){ ((n(n+1))/4)−((2n+1)/6)} =((n(n+1))/2){((3n(n+1)−2(2n+1))/(12))} =((n(n+1)(3n^2 −n−2))/(24))$
	$\sum_{1 ⩽ i < j ⩽ n} ij = \sum_{j = 1}^{n} (\sum_{i = 1}^{j - 1} ij) = \sum_{j = 1}^{n} j (\frac{(j - 1) j}{2}) = \sum_{j = 1}^{n} \frac{j^{2} (j - 1)}{2}$ $= \frac{1}{2} (\sum_{j = 1}^{n} j^{3} - \sum_{j = 1}^{n} j^{2}) = \frac{1}{2} {\frac{n^{2} {(n + 1)}^{2}}{4} - \frac{n (n + 1) (2 n + 1)}{6}}$ $= \frac{n (n + 1)}{2} {\frac{n (n + 1)}{4} - \frac{2 n + 1}{6}} = \frac{n (n + 1)}{2} {\frac{3 n (n + 1) - 2 (2 n + 1)}{12}}$ $= \frac{n (n + 1) ({3 n}^{2} - n - 2)}{24}$
	Commented by I want to learn more last updated on 17/Jun/20

	$Thanks sir$
	Commented by mathmax by abdo last updated on 17/Jun/20

	$you are welcome$
	Answered by mathmax by abdo last updated on 16/Jun/20
	$another method we have (Σ_(i=1) ^n i)^2 =Σ_(i=1) ^n i^2 +2Σ_(1≤i<j≤n) ij ⇒ Σ_(1≤i<j≤n) ij =(1/2){ (Σ_(i=1) ^n i)^2 −Σ_(i=1) ^n i^2 } =(1/2){ (((n(n+1))/2))^2 −((n(n+1)(2n+1))/6)} =(1/2){ ((n^2 (n+1)^2 )/4)−((n(n+1)(2n+1))/6)}$
	$another method we have {(\sum_{i = 1}^{n} i)}^{2} = \sum_{i = 1}^{n} i^{2} + 2 \sum_{1 ⩽ i < j ⩽ n} ij \Rightarrow$ $\sum_{1 ⩽ i < j ⩽ n} ij = \frac{1}{2} {{(\sum_{i = 1}^{n} i)}^{2} - \sum_{i = 1}^{n} i^{2}}$ $= \frac{1}{2} {{(\frac{n (n + 1)}{2})}^{2} - \frac{n (n + 1) (2 n + 1)}{6}} = \frac{1}{2} {\frac{n^{2} {(n + 1)}^{2}}{4} - \frac{n (n + 1) (2 n + 1)}{6}}$
	Commented by I want to learn more last updated on 17/Jun/20

	$Thanks sir$
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