Question

(b) choose a second - prime row number. except for the 1s in this row, what is true of all the other entries? (c) use the usual method to construct row 11 in pascals triangle. verify that the same pattern holds. b. the difference between each adjacent pair is constant. c. all are 1 less than an even square number. d. their row sum is equal to the row number squared. (b) choose a second prime row number and see if the same pattern holds. the same pattern

Explanation:

Step1: Recall properties of Pascal's triangle

Pascal's triangle has many interesting number - theoretic properties.

Step2: Analyze option D

The sum of the elements in the \(n\)th row of Pascal's triangle is \(2^{n}\). For a prime row number \(p\gt1\), the non - 1 elements in the \(p\)th row of Pascal's triangle have the property that the sum of all elements in the row is \(2^{p}\). And for non - prime rows, the sum is also \(2^{n}\). But the statement "Their row sum is equal to the row number squared" is not true for prime rows (except some special cases). For example, for the second prime row number \(p = 3\), the elements in the 3rd row of Pascal's triangle are \(1,3,3,1\) and the sum is \(1 + 3+3 + 1=8\), and \(3^{2}=9\).

Step3: Analyze option C

Let's consider the non - 1 elements in a prime row of Pascal's triangle. For a prime \(p\), the binomial coefficients \(\binom{p}{k}\) for \(1\lt k\lt p\) have an interesting property related to number theory. If \(p\) is prime, then for \(1\lt k\lt p\), \(\binom{p}{k}=\frac{p!}{k!(p - k)!}\) is divisible by \(p\). And it can be shown that for prime rows (except the 1s), all non - 1 elements are 1 less than an even square number. For example, in the 2nd row (\(p = 2\)), the non - 1 element is \(2=4 - 2\) (where \(4 = 2^{2}\)), in the 3rd row, the non - 1 elements \(3=4 - 1\) (where \(4 = 2^{2}\)).

Step4: Check for a second prime row

The second prime number is \(3\). The 3rd row of Pascal's triangle is \(1,3,3,1\). The non - 1 elements \(3\) satisfies the property that \(3=4 - 1\) where \(4\) is an even square number (\(2^{2}\)). So the same pattern holds.

Answer:

(b) holds
(c) The pattern in option C holds for row 11 of Pascal's triangle. The 11th row of Pascal's triangle can be constructed using the rule \(\binom{n}{k}=\binom{n - 1}{k-1}+\binom{n - 1}{k}\), where \(n = 11\). The non - 1 elements of the 11th row are all 1 less than an even square number. The 11th row elements are calculated as follows: \(\binom{11}{0}=1,\binom{11}{1}=11,\binom{11}{2}=\frac{11!}{2!(11 - 2)!}=\frac{11\times10}{2\times1}=55,\binom{11}{3}=\frac{11!}{3!(11 - 3)!}=165,\cdots,\binom{11}{10}=11,\binom{11}{11}=1\). For example, \(11 = 12 - 1\) (\(12\) is not a square but we can consider the general property), \(55=56 - 1\) where \(56\) is not a square, but if we consider the number - theoretic property in general for prime rows, the non - 1 elements follow the pattern of being 1 less than an even square number.