Finding any row of Pascal's triangle extending the concept of the power of 11
aa r X i v : . [ m a t h . HO ] M a y Finding any row of Pascal’s triangleextending the concept of power of Md. Robiul Islam ∗ ,1 and Md. Shorif Hossan Department of Computer Science and Engineering, Green University ofBangladesh, Dhaka, Bangladesh Department of Applied Mathematics, University of Dhaka, Dhaka,Bangladesh
E-mail: [email protected] , [email protected] ORCID: , Abstract
Generalization of Pascal’s triangle and its fascinating properties at-tract the attention of many researchers from the very beginning. SirIsaac Newton first observed that the first five rows of Pascal’s trian-gle can be obtained from the power of 11 and claimed without proofthat the subsequent rows of Pascal’s triangle can also be generated bythe power of eleven. The allegation later proved by Arnold but thevisualization of the rows restricted till 5 th row due to the limitation of11. In the concept of 11, Morton showed that dividing each row (con-sidering multi-digit numerals as single place value) by 11 we get animmediately preceding row, but he didn’t give any formula for gettingthe full row. In this paper, a formula is derived as an extension of theconcept of 11 n to generate any row of Pascal’s triangle. We extendedthe concept of 11 n to 1 Θ n . We briefly discussed how our proposedconcept works for any number of n by employing an appropriate num-ber of zeros between 1 and 1 (11) represented by Θ . We generatedthe formula for getting the value of Θ stands for the number of zerosbetween 1 and 1. The evaluation of our proposed concept verified withPascal’s triangle and matched successfully. Finally, we demonstratePascal’s triangle for a large n such as 51 st row as an example usingour proposed formula. Keywords:
Pascal’s triangle, power of 11, Finding any rows, generalizedPascal’s triangle. 1
Introduction
Algebra is a spacious part of the science of mathematics which provides theopportunity to express mathematical ideas more precisely. In algebra, theBinomial expansion and Pascal’s triangle are considered important. Pascal’striangle is a triangular arrangement of the binomial coefficients and one ofthe most known integer models. Though it was named after French scientistBlaise Pascal, it was studied in ancient India, Persia, China, Germany, andItaly by different mathematicians afore him. In fact, the definition of thetriangle was made centuries ago. It is thought that in 450 BC, Indian math-ematician Pingala was included the concept of this triangle in the book ofpoetry in Sanskrit. At the same time, the commentators of this book acquaintthat the diagonal surface of the triangle is the sum of the Fibonacci num-bers. It is the same idea among Chinese mathematicians and calls the triangle“Yang Hui’s triangle”. Later, Persian mathematician Al-Karaji and Persianastronomer-poet Omar Khayyam named the triangle as the “Khayyam tri-angle”. It also has multidimensional shapes, the three-dimensional shapeis referred to as Pascal’s pyramid or Pascal’s tetrahedron, while the othergeneral-shaped ones are called Pascal’s simplifications. Mathematicians findthe application of this triangle in mathematics and many modern physicssubjects. Various studies have been conducted in many different disciplinesabout Pascal’s triangle. The studies conducted in the last century can beanalyzed as follows.In [1], the importance of the Pascal’s triangle in modern mathematics andproperties of this triangle with an application are discussed. In [2], appli-cations on Pascal’s triangle using modular arithmetic are showed. In eachapplication, the first number was increased by one, and correlated the re-sults with the Pascal triangle. Pascal method is narrated in [3] as "the usualmethod of selection for middle school or higher level students, which deter-mines the number of a number of subsets". Here Sgroi mentioned that in theconstruction of Pascal’s triangle, each line starts with 1 and ends with 1, andthis series can be expanded with simple cross-joints. In his study, Jansson[4] developed three geometric forms related to Pascal’s triangle and includedexamples on each form. In [5], 17 different properties of Pascal’s triangle andtheir relations with each other are studied. The relationship between Pas-cal’s triangle and Binomial expansion are investigated by using permutations[6]. In his study, Toschi [7] constructed new types of Pascal’s triangles using2ifferent permutations and created generalizations. Duncan and Litwiller [8]discussed about the reconstruction of Pascal’s triangle with the individu-als. Here they collected data on the opinions of individuals using qualitativemethods, and determined the methods of constructing the Pascal’s trianglein different ways with the attained findings. The relationship between thePascal triangle and the Fibonacci numbers had been discussed In [9]. In [10],the Pascal pyramid concept created and visualized the Pascal triangle. Inhis study, Putz [11] developed the concept of Pascal Polytope using the con-cept of permutation and associated it with the Fibonacci concept. In [12],Houghton gave concept about the relationship between successive differentialoperation of a function and Pascal’s triangle. Here, he tried to integrate theconcept of differentiable function into Pascal’s triangle with an application.In [13], relationship between Pascal’s triangle and Tower of Hanoi had beenexpressed. While forming this relationship, he benefited from the Kummer’stheorem. In his study, Osler [14] affirmed that Pascal’s triangle is the oldestand most important tool in mathematics. In addition, he used it in brackets,square brackets and higher forces, and identified each of these expansionswith Pascal’s triangle.In 1956, Freund [15] elicited that the generalized Pascal’s triangles of s th or-der can be constructed from the generalized binomial coefficients of order s .In [16] Bankier gave the Freud’s alternative proof. Kallós tried to generalizePascal triangle using power of integers [17], different based triangle [18] andtheir connections with prime number [19]. Kuhlmann tried to generate Pas-cal’s triangle using the T-triangle concept [20]. Some fascinating propertiesof Pascal’s triangle are available in [21, 22].The concept of power of 11 was first introduced by sir Issac Newton. Heobserved that first five rows of Pascal’s triangle are generated by power of 11and claimed (without proof) for the later rows, that is successive rows canalso be generated by power of eleven [23]. In [24] Arnold et al. supportedNewton’s assertion and proved it generally. In [25] Mueller noted that fromthe n th row of the Pascal’s triangle with positional addition, one can get the n th power of 11. In this study, we try to extend the concept of power of 11and proposed a formula to attain any row of Pascal’s triangle.3 Methods
The very basic definition to get any element of a row of the Pascal’s triangleis the summation of two adjacent elements of the previous row. Each numberin Pascal’s triangle is the sum of two numbers above that number. Usually,the lines of Pascal’s triangle are numbered starting from n = 0 from the topand the numbers in each line are starting from k = 0 from the left. For k=0their is only one value 1. As the next lines are created, The remaining rightmost and left most element for new row is taken as 1.11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 1 . . . . . . . . .
Figure 1:
Pascal’s triangleThe concept of power of 11 leads to us 11 =
11, 1 st row of Pascal trian-gle and so 11 = = = nd , 3 rd and 4 th row respectively. Before finding the general rule for subsequent rows we firstelaborate the previous concept of 11. The reason behind getting Pascal’striangle by the power of 11 lies on the general rule of multiplication. Whatdo we get from multiplication of a number by 11? multiplication of 11 leadsto us the following figure.2 nd row of Pascal’s triangle → × → rd row of Pascal’s triangle → O Figure 2:
Results after multiplication by 114igure 2 shows, multiplication of a number by 11 gives an output which issimilar to the addition of the two adjacent numbers of previous row of Pas-cal’s triangle.Patently 11 = = th and 6 th row of Pascal’striangle are 1 5 10 10 5 1&1 6 15 20 15 6 1respectively. The above scheme fails for 11 or 11 . Why are we not gettingthe 5 th row or why does the power of 11 fail here? The answer is the middlevalue from the 5 th row of Pascal’s triangle are of two decimal places whereasthe power of 11 represents Pascal’s row as a representation of one decimalplace. So for finding 5 th or any frontal row, we need a formula that can rep-resent the number generated from the power of 11 as two or higher decimalplaces. Now, we will endeavor to formulate a specific rule that generates therequired number of decimal places for the representation of Pascal’s triangle.At first, we attempt to represent the number as two decimal places using thevery basic rules of multiplication. Figure 3, displays the impact of multipli-cation by 101 101 × → → O × → O O
01 Numbers given in-side the circles aresame as the summa-tion of two adjacentnumbers of the pre-vious row, but mul-tiplication by 101displays the rows asa representation oftwo decimal places
Figure 3:
Results after multiplication by 101Now, 101 = th row of Pascal’striangle by omitting extra zeros and separating the digits.5 5 10 10 5 1Similarly from 101 = = th and 7 th row respectively.1 06 15 20 15 06 01&1 07 21 35 35 21 07 01101 , 101 and 101 all are representing 5 th , 6 th and 7 th row of Pascal’striangle respectively as a representation of two decimal places due to theaddition of one zero between 1 and 1 (11) such that 101. 11 , 11 and 11 could also represent the respective rows according to the Newton’s claim but101 n makes the visualization.Can a conclusion be drawn for the generating any row of Pascal’s trianglewith the help of extended concept of power of 11 such as 101 n ? Let’s havea look for n=9. Plainly, 101 = th row ofPascal’s triangle is 1 9 36 84 126 126 84 36 9 1This is due to the three digits in the central element 126. So, we need aformula for the representation of three decimal places. The previous contextdirected that multiplication of a number by 11 and 101 makes the left shiftof all digits by one and two times respectively. So the representation of threedecimal places requires multiplication by 1001.Figure 4, proofs the left shift of all digits by 3 times when a number ismultiplied by 1001 6001 × → O × → O O
001 Numbers given insidethe circles are summa-tion of the two adjacentnumbers of the previousrow as a representationof three decimal places
Figure 4:
Results after multiplication by 1001By continuing the multiplication, we get1001 = th row of Pascal’s triangle with the represen-tation as three decimal places:1 009 036 084 126 126 084 036 009 001Similarly, 1001 represents the 10 th row of Pascal triangle with the repre-sentation as three decimal places:1010045120210252210120045010001 From the above study, it can be easily concluded that the representation ofthree decimal places requires the left shift of all digits by three times, andthree times the left shift of all digits requires two zeros between 1 and 1 (11),that is 1001. Why do we require three decimal places representation for 9 th and 10 th rows of Pascal’s triangle?. Because the central element of 9 th and70 th row is of three decimal places. Similarly, we required two decimal placesrepresentation for 5 th to 8 th rows since the central element of these rows arenumbers of two decimal places. And, the 1 st four rows satisfied 11 n sincethose are numbers of one decimal place. So for any row, the number of dec-imal places representation should be equal to the number of digits exist inthe central value of that row.Now we seek to generate a formula to find the central value of any rowof the Pascal’s triangle. For an odd number, say n = n + = th row. So the central value should be (cid:16) (cid:17) th = th obser-vation of that row, which is (cid:16) − (cid:17) = (cid:16) (cid:17) = n =
10 we get n + =
11 elements and the central value should be (cid:16) (cid:17) = . ⇒ th Ceiling value observation, which is (cid:16) − (cid:17) = (cid:16) (cid:17) = floor value of n .So the formula for having central value of n th row is (cid:16) nfloor n (cid:17) . But we neverneed for the central value rather get the number of digits to exist in thecentral value. Let’s make it more facilitate, using Logarithmic function wecan directly calculate how many digits (or decimal places) should the centralnumber have?. Applying the property of Logarithmic function the formulabecomes log (cid:16) nfloor n (cid:17) , since ceil log X represents the number of digits of X . However, if the central value is of d decimal places then we require oneless number of zeros between 1 and 1 (11) such that (cid:16) (d-1) zeros (cid:17) n . So,we can get the number of zeros required between 1 and 1 (11) by taking the floor value of log (cid:16) nfloor n (cid:17) that is f loor log (cid:16) nfloor n (cid:17) .Let us consider Θ represents the number of zeros between 1 and 1 (11).Then Θ = f loor (cid:18) log (cid:16) nfloor n (cid:17)(cid:19) . Let’s verify it for an odd number n = n = n = Θ = f loor log f loor !! = f loor log !! = f loor (cid:0) . (cid:1) = n =
10 gives Θ = f loor log f loor !! = f loor log !! = f loor (cid:0) . (cid:1) = th and 10 th rows we have to calculate 1001 and 1001 respectively. Bothare verified above already.It’s time to generate the formula to find any row of Pascal’s triangle. Thegeneral formula for generating n th row of Pascal’s triangle is 1 Θ n , where Θ represents the number zeros required to generate the desired row and definedby Θ = f loor log nf loor n !! For a random number such as n =
15 we get Θ =
3. So we have to insert 3zeros and the 15 th row can be constructed from the following10001 = n = Θ =
4. So the 16 th row can be constructed from the following100001 = n . We now exemplify 51 st row of Pascal’s triangle.9ence n =
51 gives Θ = f loor log f loor !! = f loor log f loor . !! = f loor log !! = f loor log (cid:0) (cid:1) = f loor . = .Now 1000000000000001 = st row can be obtained by separating each 15 digits (except thefirst digit 1) from the above result. For readers convenient, we marked eachentry with different colors and showing that the above formula generates aPascal’s triangle with a representation of 15 digits. Pascal’s triangle is a startling mathematical tool that has vastly inflictionthroughout various mathematical topics. So, forming pascal’s triangle easilyand quickly is an expectation of all analysts who are interested in it. Here,10 e extended the existing formula from 11 n to 1 Θ n . In view of the abovediscussion, we may conclude that, as multiplication by 11 leads us to theaddition of the adjacent numbers of the previous row so we can find any rowof Pascal’s triangle by inserting proper number of zeros between 1 and 1(11).The number of zeros yields from: Θ = f loor (cid:18) log (cid:16) nfloor n (cid:17)(cid:19) and the n th row is obtained by 1 Θ n . References [1] Donald E White. An approach to modern mathematics through pascal’striangle.
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