Please-explain-1-i-lt-j-n-ij-j-2-n-j-j-1-j-2-I-want-to-know-how-L-H-S-R-H-S-

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}} \dots = \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

\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 {(\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