Question

the “triangle” that pascal studied and published in his treatise was actually more like a truncated corner of tartaglias rectangle. each number in tartaglias rectangle can be calculated in various ways. consider the number n to be located anywhere in the array. by checking several locations in the given array, determine how n is related to the sum of all entries in the shaded cells.

click the icon to view the truncated corner of tartaglias rectangle.

complete the following table of the value of n and the sum of the values of the shaded cells for various locations of n in the truncated corner of tartaglias rectangle.
location of n value of n sum of values of shaded cells
row 3, column 4 10 10
row 5, column 3

Explanation:

Step1: Recall the property of Tartaglia's rectangle

In Tartaglia's rectangle - like structure (related to Pascal - like numbers), the number at a particular location is equal to the sum of the values in the shaded cells above and to the left of it.

Step2: Calculate for Row 5, Column 3

We know that the number in a cell of this structure is the sum of the shaded - cell values related to it. In a Pascal - like triangle (or this truncated corner version), we can calculate the number in the 5th row and 3rd column.
The binomial coefficient formula for the number in the $n$th row and $k$th column of Pascal's triangle is $C(n - 1,k - 1)=\frac{(n - 1)!}{(k - 1)!(n - k)!}$. Here $n = 5$ and $k=3$.
First, calculate the binomial coefficient:
\[

$$\begin{align*} C(5 - 1,3 - 1)&=C(4,2)\\ &=\frac{4!}{2!(4 - 2)!}\\ &=\frac{4!}{2!2!}\\ &=\frac{4\times3\times2!}{2!×2!}\\ &=\frac{4\times3}{2\times 1}\\ & = 6 \end{align*}$$

\]
Since the number $N$ at a location is equal to the sum of the values of the shaded cells, the sum of the values of the shaded cells is also 6.

Answer:

Row 5, Column 3: Value of N: 6, Sum of values of shaded cells: 6