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the \triangle\ that pascal studied and published in his treatise was ac…

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 8, column 2 8 8
row 6, column 3 21 21
row 4, column 4

Explanation:

Step1: Recall the property of Tartaglia's rectangle

Tartaglia's rectangle has a property where the number $N$ in a cell is equal to the sum of the values in the shaded - cells above and to the left of it.

Step2: Use combination formula for Tartaglia's rectangle

The numbers in Tartaglia's rectangle are binomial coefficients. The element in the $n$ - th row and $k$ - th column (where rows and columns start from 1) of Tartaglia's rectangle is given by the binomial coefficient $\binom{n - 1}{k - 1}$.
For row $n = 4$ and column $k = 4$, we calculate the binomial coefficient $\binom{n-1}{k - 1}=\binom{4 - 1}{4 - 1}=\binom{3}{3}$.
We know that the binomial coefficient formula is $\binom{n}{r}=\frac{n!}{r!(n - r)!}$.
So, $\binom{3}{3}=\frac{3!}{3!(3 - 3)!}=\frac{3!}{3!0!}=1$.
Since the number $N$ is equal to the sum of the values in the shaded cells, the sum of the values of the shaded cells is also 1.

Answer:

Row 4, Column 4: Value of N: 1, Sum of values of shaded cells: 1