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" Let " Delta(r)=|{:(r-1,,n,,6),((r-1)^...

`" Let " Delta_(r)=|{:(r-1,,n,,6),((r-1)^(2),,2n^(2),,4n-2),((r-1)^(2),,3n^(3),,3n^(2)-3n):}|. " Show that " Sigma_(r=1)^(n) Delta_(r)`
is constant.

Text Solution

Verified by Experts

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 "]`
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