The value of left|begin{array}{lll}(a+1)(a+2) | vedclass

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(B) Let $\Delta = \left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \ (a+2)(a+3) & a+3 & 1 \ (a+3)(a+4) & a+4 & 1\end{array}ight|$.Applying row operati

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$-2$

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(B) Let $\Delta = \left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \ (a+2)(a+3) & a+3 & 1 \ (a+3)(a+4) & a+4 & 1\end{array}ight|$.Applying row operations $R_{2} ightarrow R_{2} - R_{1}$ and $R_{3} ightarrow R_{3} - R_{1}$:$\Delta = \left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \ (a+2)(a+3) - (a+1)(a+2) & (a+3) - (a+2) & 1 - 1 \ (a+3)(a+4) - (a+1)(a+2) & (a+4) - (a+2) & 1 - 1\end{array}ight|$$\Delta = \left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \ (a+2)(a+3-a-1) & 1 & 0 \ (a^2+7a+12) - (a^2+3a+2) & 2 & 0\end{array}ight|$$\Delta = \left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \ 2(a+2) & 1 & 0 \ 4a+10 & 2 & 0\end{array}ight|$Expanding along the third column:$\Delta = 1 	imes [2(a+2) 	imes 2 - 1 	imes (4a+10)]$$\Delta = 4(a+2) - (4a+10)$$\Delta = 4a + 8 - 4a - 10$$\Delta = -2$.

The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ is

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