In mathematics, Pascal's triangle is an infinite triangular array of the binomial coefficients which play a crucial role in probability theory, combinatorics, and algebra. In much of the Western world, it is named after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in Persia, India, China, Germany, and Italy. The rows of Pascal's triangle are conventionally enumerated starting with row displaystylen=0 at the top (the 0th row). The entries in each row are numbered from the left beginning with displaystylek=0 and are usually staggered relative to the numbers in the adjacent rows. The triangle may be constructed in the following manner: In row 0 (the topmost row), there is a unique nonzero entry 1. Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right, treating blank entries as 0. For example, the initial number of row 1 (or any other row) is 1 (the sum of 0 and 1), whereas the numbers 1 and 3 in row 3 are added to produce the number 4 in row 4. In the displaystyle nth row of Pascal's triangle, the displaystyle kth entry is denoted displaystyletbinomnk, pronounced "n choose k". For example, the topmost entry is displaystyletbinom00=1. With this notation, the construction of the previous paragraph may be written as displaystylen choose k=n-1choosek-1+n-1choose k for any positive integer displaystyle n and any integer displaystyle0leq kleq n. This recurrence for the binomial coefficients is known as Pascal's rule. The pattern of numbers that forms Pascal's triangle was known well before Pascal's time. The Persian mathematician Al-Karaji wrote a now-lost book which contained the first formulation of the binomial coefficients and the first description of Pascal's triangle. It was later repeated by Omar Khayyám, another Persian mathematician; thus the triangle is also referred to as Khayyam's triangle in Iran. Several theorems related to the triangle were known, including the binomial theorem. Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients. Pascal's triangle was known in China during the early 11th century through the work of the Chinese mathematician Jia Xian. During the 13th century, Yang Hui defined the triangle, and it is known as Yang Hui's triangle in China. In Europe, Pascal's triangle appeared for the first time in the Arithmetic of Jordanus de Nemore. The binomial coefficients were calculated by Gersonides during the early 14th century, using the multiplicative formula for them. Petrus Apianus published the full triangle on the frontispiece of his book on business calculations in 1527. Michael Stifel published a portion of the triangle in 1544, describing it as a table of figurate numbers. In Italy, Pascal's triangle is referred to as Tartaglia's triangle, named for the Italian algebraist Tartaglia, who published six rows of the triangle in 1556. Gerolamo Cardano also published the triangle as well as the additive and multiplicative rules for constructing it in 1570. Pascal's Traité du triangle arithmétique was published posthumously in 1665. In this, Pascal collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named for Pascal by Pierre Raymond de Montmort who called it table de M. Pascal pour les combinaisons and Abraham de Moivre who called it Triangulum Arithmeticum PASCALIANUM, which became the basis of the modern Western name. Pascal's triangle determines the coefficients which arise in binomial expansions. For example, in the expansion displaystyle(x+y)²=x²+2xy+y²=mathbf1x²y⁰+mathbf2x¹y¹+mathbf1x⁰y², the coefficients are the entries in the second row of Pascal's triangle: displaystyletbinom20=1, displaystyletbinom21=2, displaystyletbinom22=1. In general, the binomial theorem states that when a binomial like displaystylex+y is raised to a positive integer power displaystyle n, the expression expands as displaystyle(x+y)ⁿ=sumₖ₌₀ⁿaₖxⁿ⁻ᵏyᵏ=a₀xⁿ+a₁xⁿ⁻¹y+a₂xⁿ⁻²y²+ldots+aₙ₋₁xyⁿ⁻¹+aₙyⁿ, where the coefficients displaystyleaₖ are precisely the numbers in row displaystyle n of Pascal's triangle: displaystyleaₖ=n choose k. The entire left diagonal of Pascal's triangle corresponds to the coefficient of displaystylexⁿ in these binomial expansions, while the next left diagonal corresponds to the coefficient of displaystylexⁿ⁻¹y, and so on. To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of displaystyle(x+y)ⁿ⁺¹ in terms of the corresponding coefficients of displaystyle(x+1)ⁿ, where we set displaystyley=1 for simplicity. Suppose then that displaystyle(x+1)ⁿ=sumₖ₌₀ⁿaₖxᵏ. Now displaystyle(x+1)ⁿ⁺¹=(x+1)(x+1)ⁿ=x(x+1)ⁿ+(x+1)ⁿ=sumᵢ₌₀ⁿaᵢxⁱ⁺¹+sumₖ₌₀ⁿaₖxᵏ. The two summations can be reindexed with displaystylek=i+1 and combined to yield displaystylebeginalignedsumᵢ₌₀ⁿaᵢxⁱ⁺¹+sumₖ₌₀ⁿaₖxᵏ&=sumₖ₌₁ⁿ⁺¹aₖ₋₁xᵏ+sumₖ₌₀ⁿaₖxᵏ[4pt]&=sumₖ₌₁ⁿaₖ₋₁xᵏ+aₙxⁿ⁺¹+a₀x⁰+sumₖ₌₁ⁿaₖxᵏ[4pt]&=a₀x⁰+sumₖ₌₁ⁿ(aₖ₋₁+aₖ)xᵏ+aₙxⁿ⁺¹[4pt]&=x⁰+sumₖ₌₁ⁿ(aₖ₋₁+aₖ)xᵏ+xⁿ⁺¹.endaligned Thus the extreme left and right coefficients remain as 1, and for any given displaystyle0<k<n+1, the coefficient of the displaystylexᵏ term in the polynomial displaystyle(x+1)ⁿ⁺¹ is equal to displaystyleaₖ₋₁+aₖ, the sum of the displaystylexᵏ⁻¹ and displaystylexᵏ coefficients in the previous power displaystyle(x+1)ⁿ. This is indeed the downward-addition rule for constructing Pascal's triangle. It is not difficult to turn this argument into a proof of the binomial theorem. Since displaystyle(a+b)ⁿ=bⁿ(tfracab+1)ⁿ, the coefficients are identical in the expansion of the general case. An interesting consequence of the binomial theorem is obtained by setting both variables displaystylex=y=1, so that displaystylesumₖ₌₀ⁿn choose k=nchoose0+nchoose1+cdots+nchoosen-1+n choose n=(1+1)ⁿ=2ⁿ. In other words, the sum of the entries in the displaystyle nth row of Pascal's triangle is the displaystyle nth power of 2. This is equivalent to the statement that the number of subsets of an displaystyle n-element set is displaystyle2ⁿ, as can be seen by observing that each of the displaystyle n elements may be independently included or excluded from a given subset. A second useful application of Pascal's triangle is in the calculation of combinations. The number of combinations of displaystyle n items taken displaystyle k at a time, i.e. the number of subsets of displaystyle k elements from among displaystyle n elements, can be found by the equation displaystylemathbfC(n,k)=mathbfCₖⁿ=ₙCₖ=n choose k=fracn!k!(n-k)!. This is equal to entry displaystyle k in row of displaystyle n Pascal's triangle. Rather than performing the multiplicative calculation, one can simply look up the appropriate entry in the triangle (constructed by additions). For example, suppose 3 workers need to be hired from among 7 candidates; then the number of possible hiring choices is 7 choose 3, the entry 3 in row 7 of the above table, which is displaystyletbinom73=35. When divided by displaystyle2ⁿ, the displaystyle nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where displaystylep=tfrac12. By the central limit theorem, this distribution approaches the normal distribution as displaystyle n increases. This can also be seen by applying Stirling's formula to the factorials involved in the formula for combinations. This is related to the operation of discrete convolution in two ways. First, polynomial multiplication corresponds exactly to discrete convolution, so that repeatedly convolving the sequence displaystyleldots,0,0,1,1,0,0,ldots with itself corresponds to taking powers of displaystylex+1, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a random variable with itself corresponds to calculating the distribution function for a sum of n independent copies of that variable; this is exactly the situation to which the central limit theorem applies, and hence results in the normal distribution in the limit. (The operation of repeatedly taking a convolution of something with itself is called the convolution power.) Pascal's triangle has many properties and contains many patterns of numbers. The sum of the elements of a single row is twice the sum of the row preceding it. For example, row 0 (the topmost row) has a value of 1, row 1 has a value of 2, row 2 has a value of 4, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row displaystyle n equals to displaystyle2ⁿ. Taking the product of the elements in each row, the sequence of products (sequence A001142 in the OEIS) is related to the base of the natural logarithm, e. Specifically, define the sequence displaystylesₙ for all displaystylengeq0 as follows: displaystylesₙ=prodₖ₌₀ⁿn choose k=prodₖ₌₀ⁿfracn!k!(n-k)! Then, the ratio of successive row products is displaystylefracsₙ₊₁sₙ=fracdisplaystyle(n+1)!ⁿ⁺²prodₖ₌₀ⁿ⁺¹frac1k!²displaystylen!ⁿ⁺¹prodₖ₌₀ⁿfrac1k!²=frac(n+1)ⁿn! and the ratio of these ratios is displaystylefracsₙ₊₁cdotsₙ₋₁sₙ²=left(fracn+1nright)ⁿ, ngeq1. The right-hand side of the above equation takes the form of the limit definition of displaystyle e displaystylee=limₙₜₒᵢₙfₜyleft(1+frac1nright)ⁿ. displaystylepi can be found in Pascal's triangle by use of the Nilakantha infinite series. displaystylepi=3+sumₙ₌₁ⁱⁿᶠᵗʸ(-1)ⁿ⁺¹frac2n+1choose12n+1choose22n+2choose2 Some of the numbers in Pascal's triangle correlate to numbers in Lozanić's triangle. The sum of the squares of the elements of row n equals the middle element of row 2n. For example, 1² + 4² + 6² + 4² + 1² = 70. In general form, displaystylesumₖ₌₀ⁿn choose k²=2n choose n. In any even row displaystylen=2m, the middle term minus the term two spots to the left equals a Catalan number, specifically displaystyleCₘ₋₁=tbinom2mm-tbinom2mm-2. For example, in row 4, which is 1, 4, 6, 4, 1, we get the 3rd Catalan number displaystyleC₃=6-1=5. In a row p, where p is a prime number, all the terms in that row except the 1s are divisible by p. This can be proven easily, from the multiplicative formula displaystyletbinompk=tfracp!k!(p-k)!. Since the denominator displaystylek!(p-k)! can have no prime factors equal to p, so p remains in the numerator after integer division, making the entire entry a multiple of p. Parity: To count odd terms in row n, convert n to binary. Let x be the number of 1s in the binary representation. Then the number of odd terms will be 2. These numbers are the values in Gould's sequence. Every entry in row 2 − 1, n ≥ 0, is odd.If we consider 1st row which contains 1 as the 0th row of the Pascal's triangle, the only rows with all terms odd will be the Mersenne numbers i.e., (2^n - 1)th row. After that, repeated flanking of even terms by O, OO, OEO, OOOO, OEOEO, OOOOOO, etc. happens. If even number of terms flank the even terms in between, then the flanking terms are all odd. If odd number of terms flank the even terms in between, then the flanking terms alternate between odd and even numbers, the first and the last numbers being odd. The parity property holds differently according to diagonals. The first diagonal from right or left side contains only 1s, all odd numbers. Then the pattern happens to be alternating odd and even (all natural numbers in ascending order), alternating two odd and two even, alternating odd then three even. The OOOO, OEOE, OOEE, OEEE pattern continues (and repeats itself a natural number of times in increasing order- first 1 time, then 2 times, 3, 4, 5, etc.) after that but consists of a multiple of four number of even terms after that. First time, 4 even numbers follow, then 8, then 12, etc. The multiples of four come in ascending order. Polarity: When the elements of a row of Pascal's triangle are alternately added and subtracted together, the result is 0. For example, row 6 is 1, 6, 15, 20, 15, 6, 1, so the formula is 1 − 6 + 15 − 20 + 15 − 6 + 1 = 0. The diagonals of Pascal's triangle contain the figurate numbers of simplices: The diagonals going along the left and right edges contain only 1's. The diagonals next to the edge diagonals contain the natural numbers in order. The 1-dimensional simplex numbers increment by 1 as the line segments extend to the next whole number along the number line. Moving inwards, the next pair of diagonals contain the triangular numbers in order. The next pair of diagonals contain the tetrahedral numbers in order, and the next pair give pentatope numbers. displaystylebeginalignedP₀(n)&=Pd(0)=1,¶d(n)&=Pd(n-1)+Pd₋₁(n)&=sumᵢ₌₀ⁿPd₋₁(i)=sumᵢ₌₀ᵈPᵢ(n-1).endaligned The symmetry of the triangle implies that the nᵗʰ d-dimensional number is equal to the dᵗʰ n-dimensional number. An alternative formula that does not involve recursion is displaystylePd(n)=frac1d!prodₖ₌₀ᵈ⁻¹(n+k)=n⁽ᵈ⁾overd!=binomn+d-1d, where n is the rising factorial. The geometric meaning of a function Pd is: Pd = 1 for all d. Construct a d-dimensional triangle by placing additional dots below an initial dot, corresponding to Pd = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find Pd, have a total of x dots composing the target shape. Pd then equals the total number of dots in the shape. A 0-dimensional triangle is a point and a 1-dimensional triangle is simply a line, and therefore P₀ = 1 and P₁ = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to Pd − 1. There are simple algorithms to compute all the elements in a row or diagonal without computing other elements or factorials. To compute row displaystyle n with the elements displaystyletbinomn0,tbinomn1,ldots,tbinomnn, begin with displaystyletbinomn0=1. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator: displaystylen choose k=nchoosek-1timesfracn+1-kk. For example, to calculate row 5, the fractions are displaystyletfrac51, displaystyletfrac42, displaystyletfrac33, displaystyletfrac24 and displaystyletfrac15, and hence the elements are displaystyletbinom50=1, displaystyletbinom51=1timestfrac51=5, displaystyletbinom52=5timestfrac42=10, etc. (The remaining elements are most easily obtained by symmetry.) To compute the diagonal containing the elements displaystyletbinomn0,tbinomn+11,tbinomn+22,ldots, begin again with displaystyletbinomn0=1 and obtain subsequent elements by multiplication by certain fractions: displaystylen+k choose k=n+k-1choosek-1timesfracn+kk. For example, to calculate the diagonal beginning at displaystyletbinom50, the fractions are displaystyletfrac61,tfrac72,tfrac83,ldots, and the elements are displaystyletbinom50=1,tbinom61=1timestfrac61=6,tbinom72=6timestfrac72=21, etc. By symmetry, these elements are equal to displaystyletbinom55,tbinom65,tbinom75, etc. The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal known as the Sierpinski triangle. This resemblance becomes increasingly accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle, assuming a fixed perimeter. More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns. As the proportion of black numbers tends to zero with increasing n, a corollary is that the proportion of odd binomial coefficients tends to zero as n tends to infinity. In a triangular portion of a grid, the number of shortest grid paths from a given node to the top node of the triangle is the corresponding entry in Pascal's triangle. On a Plinko game board shaped like a triangle, this distribution should give the probabilities of winning the various prizes. If the rows of Pascal's triangle are left-justified, the diagonal bands sum to the Fibonacci numbers. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the matrix exponential can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4,... on its subdiagonal and zero everywhere else. Labelling the elements of each n-simplex matches the basis elements of Clifford algebra used as forms in Geometric Algebra rather than matrices. Recognising the geometric operations, such as rotations, allows the algebra operations to be discovered. Just as each row, n, starting at 0, of Pascal's triangle corresponds to an (n-1)-simplex, as described below, it also defines the number of named basis forms in n dimensional Geometric algebra. The binomial theorem can be used to prove the geometric relationship provided by Pascal's triangle. This same proof could be applied to simplices except that the first column of all 1's must be ignored whereas in the algebra these correspond to the real numbers, displaystylemathbbR, with basis 1. Pascal's triangle can be used as a lookup table for the number of elements within a polytope. Let's begin by considering the 3rd line of Pascal's triangle, with values 1, 3, 3, 1. A 2-dimensional triangle has one 2-dimensional element, three 1-dimensional elements, and three 0-dimensional elements. The meaning of the final number is more difficult to explain. Continuing with our example, a tetrahedron has one 3-dimensional element, four 2-dimensional elements, six 1-dimensional elements, and four 0-dimensional elements. Adding the final 1 again, these values correspond to the 4th row of the triangle. Line 1 corresponds to a point, and Line 2 corresponds to a line segment. This pattern continues to arbitrarily high-dimensioned hyper-tetrahedrons. To understand why this pattern exists, one must first understand that the process of building an n-simplex from an -simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices. To build a tetrahedron from a triangle, position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle. The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells is 0 + 1 = 1; the number of faces is 1 + 3 = 4; the number of edges is 3 + 3 = 6; the number of new vertices is 3 + 1 = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle. A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2)ʳᵒʷ ⁿᵘᵐᵇᵉʳ, instead of (x + 1)ʳᵒʷ ⁿᵘᵐᵇᵉʳ. There are a couple ways to do this. The simpler is to begin with row 0 = 1 and row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule: displaystylen choose k=2timesn-1choosek-1+n-1choose k. That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in: displaystylebeginmatrixtext1{text1}quadtext2{text1}quadtext4quadtext4{text1}quadtext6quadtext12quadtext8{text1}quadtext8quadtext24quadtext32quadtext16{text1}quadtext10quadtext40quadtext80quadtext80quadtext32{text1}quadtext12quadtext60quad160quad240quad192quadtext64{text1}quadtext14quadtext84quad280quad560quad672quad448quad128endmatrix The other way of producing this triangle is to start with Pascal's triangle and multiply each entry by 2ᵏ, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6. To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2ᵖᵒˢⁱᵗⁱᵒⁿ ⁿᵘᵐᵇᵉʳ = 6 × 2² = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube is one, the number of 1-dimensional elements is 4, and the number of 0-dimensional elements is 4. This matches the 2nd row of the table. A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle. This pattern continues indefinitely. To understand why this pattern exists, first recognize that the construction of an n-cube from an -cube is done by simply duplicating the original figure and displacing it some distance orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n-cube and, as before, new elements are built upon those of one fewer dimension. Again, the last number of a row represents the number of new vertices to be added to generate the next higher n-cube. In this triangle, the sum of the elements of row m is equal to 3. Again, to use the elements of row 4 as an example: 1 + 8 + 24 + 32 + 16 = 81, which is equal to displaystyle3⁴=81. Each row of Pascal's triangle gives the number of vertices at each distance from a fixed vertex in an n-dimensional cube. For example, in three dimensions, the third row corresponds to the usual three-dimensional cube: fixing a vertex V, there is one vertex at distance 0 from V, three vertices at distance 1, three vertices at distance √2 and one vertex at distance √3. The second row corresponds to a square, while larger-numbered rows correspond to hypercubes in each dimension. As stated previously, the coefficients of are the nth row of the triangle. Now the coefficients of are the same, except that the sign alternates from +1 to −1 and back again. After suitable normalization, the same pattern of numbers occurs in the Fourier transform of sin/x. More precisely: if n is even, take the real part of the transform, and if n is odd, take the imaginary part. Then the result is a step function, whose values are given by the nth row of the triangle with alternating signs. For example, the values of the step function that results from: displaystylemathfrakReleft(textFourierleft[fracsin(x)⁵xright]right) compose the 4th row of the triangle, with alternating signs. This is a generalization of the following basic result: displaystylemathfrakReleft(textFourierleft[fracsin(x)¹xright]right) is the boxcar function. The corresponding row of the triangle is row 0, which consists of just the number 1. If n is congruent to 2 or to 3 mod 4, then the signs start with −1. In fact, the sequence of the (normalized) first terms corresponds to the powers of i, which cycle around the intersection of the axes with the unit circle in the complex plane: displaystyle+i,-1,-i,+1,+i,ldots Pascal's triangle may be extended upwards, above the 1 at the apex, preserving the additive property, but there is more than one way to do so. Pascal's triangle has higher dimensional generalizations. The three-dimensional version is known as Pascal's pyramid or Pascal's tetrahedron, while the general versions are known as Pascal's simplices. When the factorial function is defined as displaystylez!=Gamma(z+1), Pascal's triangle can be extended beyond the integers to displaystylemathbbC, since displaystyleGamma(z+1) is meromorphic to the entire complex plane. Isaac Newton once observed that the first five rows of Pascal's triangle, when read as the digits of an integer, are the corresponding powers of eleven. He claimed without proof that subsequent rows also generate powers of eleven. In 1964, Robert L. Morton presented the more generalized argument that each row displaystyle n can be read as a radix displaystyle a numeral, where displaystylelimₙₜₒᵢₙfₜy11ₐⁿ is the hypothetical terminal row, or limit, of the triangle, and the rows are its partial products. He proved the entries of row displaystyle n, when interpreted directly as a place-value numeral, correspond to the binomial expansion of displaystyle(a+1)ⁿ=11ₐⁿ. More rigorous proofs have since been developed. To better understand the principle behind this interpretation, here are some things to recall about binomials: A radix displaystyle a numeral in positional notation (e.g. displaystyle14641ₐ) is a univariate polynomial in the variable displaystyle a, where the degree of the variable of the displaystyle ith term (starting with displaystylei=0) is displaystyle i. For example, displaystyle14641ₐ=1cdota⁴+4cdota³+6cdota²+4cdota¹+1cdota⁰. A row corresponds to the binomial expansion of displaystyle(a+b)ⁿ. The variable displaystyle b can be eliminated from the expansion by setting displaystyleb=1. The expansion now typifies the expanded form of a radix displaystyle a numeral, as demonstrated above. Thus, when the entries of the row are concatenated and read in radix displaystyle a they form the numerical equivalent of displaystyle(a+1)ⁿ=11ₐⁿ. If displaystylec=a+1 for displaystylec<0, then the theorem holds for displaystylea=c-1,-(c+1);mathrmmod;2c with odd values of displaystyle n yielding negative row products. By setting the row's radix (the variable displaystyle a) equal to one and ten, row displaystyle n becomes the product displaystyle11₁ⁿ=2ⁿ and displaystyle11₁₀ⁿ=11ⁿ, respectively. To illustrate, consider displaystylea=n, which yields the row product displaystylenⁿleft(1+frac1nright)ⁿ=11ₙⁿ. The numeric representation of displaystyle11ₙⁿ is formed by concatenating the entries of row displaystyle n. The twelfth row denotes the product: displaystyle11₁₂¹²=1:10:56:164:353:560:650:560:353:164:56:10:1₁₂=27433a9699701₁₂ with compound digits (delimited by ":") in radix twelve. The digits from displaystylek=n-1 through displaystylek=1 are compound because these row entries compute to values greater than or equal to twelve. To normalize the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient displaystylenchoosen-1 from its leftmost digit up to, but excluding, its rightmost digit, and use radix-twelve arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this particular example, the normalized string ends with displaystyle01 for all displaystyle n. The leftmost digit is displaystyle2 for displaystylen>2, which is obtained by carrying the displaystyle1 of displaystyle10ₙ at entry displaystylek=1. It follows that the length of the normalized value of displaystyle11ₙⁿ is equal to the row length, displaystylen+1. The integral part of displaystyle1.1ₙⁿ contains exactly one digit because displaystyle n (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of displaystyle1.1₁₂₃₄¹²³⁴. Compound digits remain in the value because they are radix displaystyle1234 residues represented in radix ten: displaystyle1.1₁₂₃₄¹²³⁴=2.885:2:35:977:696:overbraceldotsᵗᵉˣᵗ¹²²⁷ᵈⁱᵍⁱᵗˢ:0:1₁₂₃₄=2.717181235ldots₁₀ "Pascal triangle", Encyclopedia of Mathematics, EMS Press, 2001 Weisstein, Eric W. "Pascal's triangle". MathWorld.
