1. Since $a_1,a_2,\dots,a_n$ are in H.P., their reciprocals are in A.P.

   Let $\frac{1}{a_1},\frac{1}{a_2},\dots,\frac{1}{a_n}$ be an arithmetic progression with common difference $d$.

2. Then for consecutive terms, $\frac{1}{a_{k+1}}-\frac{1}{a_k}=d$ for all $k=1,2,\dots,n-1$.

   Rearranging, $\frac{a_k-a_{k+1}}{a_ka_{k+1}}=d$ so $a_ka_{k+1}=\frac{a_k-a_{k+1}}{d}.$

3. Now sum from $k=1$ to $n-1$: $\sum_{k=1}^{n-1} a_ka_{k+1}=\frac{1}{d}\sum_{k=1}^{n-1}(a_k-a_{k+1}).$

   The sum on the right telescopes: $\sum_{k=1}^{n-1}(a_k-a_{k+1})=a_1-a_n.$

   Hence, $\sum_{k=1}^{n-1} a_ka_{k+1}=\frac{a_1-a_n}{d}.$

4. We now express $d$ in terms of $a_1$ and $a_n$.

   Since the reciprocals form an A.P., $\frac{1}{a_n}=\frac{1}{a_1}+(n-1)d.$

   Therefore, $(n-1)d=\frac{1}{a_n}-\frac{1}{a_1}=\frac{a_1-a_n}{a_1a_n}.$

   So, $d=\frac{a_1-a_n}{(n-1)a_1a_n}.$

5. Substitute into the expression:

   =\frac{a_1-a_n}{\dfrac{a_1-a_n}{(n-1)a_1a_n}} =(n-1)a_1a_n.$$

6. Therefore, $a_1a_2+a_2a_3+\cdots+a_{n-1}a_n=(n-1)a_1a_n.$

So the correct option is: $\boxed{\text{D: }(n-1)a_1a_n}$