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Monoxdifly
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 Aug 6, 2015
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Help me if what I have done so far can be simplified further.
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You would do better to evaluate this determinant using row and column operations. Start by subtracting row 2 from row 3, using calculations like $(n+2)^2  (n+1)^2 = n^2+4n+4  (n^2+2n+1) = 2n+3$: $$ \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix} = \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix}.$$ Then continue like this: $$\begin{aligned} \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \end{vmatrix} &= \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix} \\ \\ \text{(Subtract row 1 from row 2)}\qquad &= \begin{vmatrix}n^2 & (n+1)^2 & (n+2)^2 \\ 2n+1 & 2n+3 & 2n+5 \\ 2n+3 & 2n+5 & 2n+7 \end{vmatrix} \\ \\ \text{(Subtract col 2 from col 3)}\qquad &= \begin{vmatrix}n^2 & (n+1)^2 & 2n+3 \\ 2n+1 & 2n+3 & 2\\ 2n+3 & 2n+5 & 2 \end{vmatrix} \\ \\ \text{(Subtract col 1 from col 2)}\qquad &= \begin{vmatrix}n^2 & 2n+1 & 2n+3 \\ 2n+1 & 2& 2\\ 2n+3 & 2 & 2 \end{vmatrix} .\end{aligned}$$ Now subtract col 2 from col 3. Proceed in this way and you should end with a very simple answer, namely the constant $8$.Help me if what I have done so far can be simplified further.