Prove that.

Let LHS $=\Delta=\mid(a+1)(a+2) \quad a+2 \quad 1$

$(a+2)(a+3) \quad a+3 \quad 1$

$(a+3)(a+4) \quad a+4 \quad 1 \mid$

$\mid(a+1)(a+2)-(a+2)(a+3) \quad(a+2)-(a+3) \quad 0(a+2)(a+3)-(a+3)(a+4) \quad(a+3)-(a+4) \quad 0(a+3)(a+4)$ $(a+4)$ $1 \mid$

[Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$ and $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$ ]

$\begin{array}{lll}=\mid-2(a+2) & -1 & 0 \\ -2(a+3) & -1 & 0\end{array}$

$\begin{aligned}&(a+3)(a+4) \quad(a+4) \quad 1 \\&=\{1 \times \mid-2(a+2)\end{aligned} \quad-1$

$-2(\mathrm{a}+3) \quad-1 \mid\} \quad\left[\right.$ Expanding along $\left.\mathrm{C}_{3}\right]$

$=4+2 \mathrm{a}-2 \mathrm{a}-6$

$=-2$

$=$ RHS

Hence proved.