This problem can be solved by Dynamic Programming, because of the property that there are overlapping subproblems.Means we are computing same thing over and again if we follow tree method.
$\color{MidnightBlue}{(i,j)} \color{blue}{\rightarrow}$ Minimum number of scalar multiplications required to find $A_{i}A_{i+1}.....A_{j}$
$A_{1} = 10*100, A_{2} = 100*5, A_{3} = 5*50$ and $A_{4} = 50*1$
Here, All unique sub-problems are $\Rightarrow$ $\color{green}{(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)}$. And we are intereseted in finding $(1,4)$ (minimum scalar multiplications to compute $A_{1}A_{2}A_{3}A_{4}$)
$\color{blue}{(1,1) = (2,2) = (3,3) = (4,4) = 0}$ (Base case)
$\color{green}{(1,2) = 10*100*5 = 5000, (2,3) = 100*5*50 = 25000}$ and $\color{green}{(3,4) = 5*50*1 = 250}$ And Dimensions of $A_{1}A_{2} , A_{2}A_{3}$ and $A_{3}A_{4}$ are $10*5, 100*50, 5*1$ respectively.
So,far table looks like :
$(1,3) = min\Bigg\{\begin{align*}
(1,1) + (2,3) &+ 10*100*50 \\
(1,2) + (3,3) &+ 10*5*50\\
\end{align*}$
$\color{olive}{(1,3) = min\Bigg\{\begin{align*}
0+ 25000 &+ 50000 \\
5000 + 0 &+ 2500\\
\end{align*} =7500}$
$(2,4) = min\Bigg\{\begin{align*}
(2,2) + (3,4) &+ 100*5*1 \\
(2,3) + (4,4) &+ 100*50*1\\
\end{align*}$
$\color{olive}{(2,4) = min\Bigg\{\begin{align*}
0 + 250&+ 500 \\
25000 + 0 &+ 5000\\
\end{align*} = 750}$
At this point, Table looks like :
We are just $1-step$ away from final result that is computing $(1,4)$ by using table above.
$\color{navy}{(1,4) = min\Bigg\{\begin{align*}
(1,1) + (2,4) &+ 10*100*1 \\
(1,2) + (3,4) &+ 10*5*1\\
(1,3) + (4,4) &+ 10*50*1
\end{align*}}$
$\color{navy}{(1,4) = min\Bigg\{\begin{align*}
0 + 750 + 1000 \\
5000 + 250 + 50\\
7500 + 0+ 500
\end{align*} = 1750}$
We got our answer to be $\color{teal}{1750}$ and Our Table looks like :
