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(C) Let the given equation be $D_1 + D_2 = 0$. First,we observe that the second determinant $D_2$ can be transposed without changing its value. $D_2 =

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any odd integer

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(C) Let the given equation be $D_1 + D_2 = 0$. First,we observe that the second determinant $D_2$ can be transposed without changing its value. $D_2 = \left| \begin{array}{ccc} a+1 & a-1 & (-1)^{n+2}a \ b+1 & b-1 & (-1)^{n+1}b \ c-1 & c+1 & (-1)^n c \end{array} ight|$. By adding the two determinants,we find that for the sum to be zero for arbitrary $a, b, c$,the columns must cancel out. Specifically,if $n$ is an odd integer,then $(-1)^{n+2} = -1$,$(-1)^{n+1} = 1$,and $(-1)^n = -1$. Substituting these into the sum of determinants,the terms cancel out to zero. Thus,$n$ must be any odd integer.

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Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is:

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