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Prove that sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)

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Your problem `->` Prove that sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)


L.H.S. `= sum [ sum (2n -3) ]`

`= sum [ sum 2n - sum 3 ]`

`= sum [ 2 × n/2 (n + 1) -3n ]`

`= sum [ n (n + 1) - 3n ]`

`= sum [ n^2 + n - 3n ]`

`= sum [ n^2 - 2n ]`

`= sum n^2 - 2 sum n`

`= (n (n + 1) (2n + 1))/6 - 2 * (n (n + 1))/2`

`= (n (n + 1) [ (2n + 1) - 6 ])/6`

`= (n (n + 1) (2n - 5))/6`

`=` R.H.S. (Proved)






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