Let `"Delta"_r=|r-1n6(r-1)^2 2n^2 4n-2(r-1)^2 3n^3 3n^2-3n|dot` Show that `sum_(r=1)^n"Delta"_r` is contant.

Explanation

Since `c_(1)` has variable terms and `c_(2) " and " c_(3)` have constant terms summation is taken to `C_(1)` Therefore,
`overset(n)underset(r=1)(Sigma) |{:(overset(n)underset(1)(Sigma)(r-1),,n,,6),(overset(n)underset(1)(Sigma)(r-1)^(2),,2n^(2),,4n-2),(overset(n)underset(1)(Sigma)(r-1)^(3),,3n^(3),,3n^(2)-3n):}|`
`|{:((1)/(2)(n-1)n,,n,,6),((1)/(6) (n-1)(2n-1),,2n^(2),,4n-2),((1)/(4) (n-1)^(2)n^(2),,3n^(3),,3n^(2)-3n):}|`
Taking `(1)/(12) n(n-1)` common from `C_(1) " and " n` common from `C_(2)` we get
` Sigma Delta_(r)=(1)/(12)n^(2)(n-1) xx|{:(6,,1,,6),(2(2n-1),,2n,,2(2n-1)),(3n(n-1),,3n^(2),,3n(n-1)):}|`
`=0 " Which is constant " [:. C_(1) " and " C_(3) " are identical "]`