**Inverse Leibniz Triangle
**

If one takes the Pascal triangle and to
the elements of |

zeroeth row , multiplies 1, |

1st row, multiplies 2 |

2nd row, multiplies 3 |

............. |

nth row, multiplies (n+1) |

One gets the Inverse Leibniz triangle. It is so called because if
one divides 1 by each element of the triangle, corresponding elements of
Leibniz triangle emerge. For example, in the 1st row, the elements are
inverted , we get 1/2,1/2; in the second row, we get 1/3, 1/6, 1/3 and
so on so forth. |

With mth element of nth row of Pascal triangle being C(n,m), the corresponding element in Leibniz triangle is the 1/ (n+1)C(n,m). Every entry is the sum of 2 numbers just below it. The entries thus can be computed sequentially left to right and top to bottom using subtraction in stead of addition. Thus, in 5th row 1/20 = (1/4) - (1/5) ; 1/30= (1/12) - (1/20) etc. |

Beauty of Pascal's Triangle |

* 1st row contains only 1s. * 2nd row contains counting numbers. * 3rd row contains triangular numbers. * 4th row contains tetrahedral numbers. * 5th row contains pentatope numbers. These are numbers which exist in 4-D space and describe the no. of vertices in a configuration of 3 dimensional tetrahedrons joined at the faces. * sum of binomial reciprocals : Discovery of π in Pascal's triangle can be attributed to Daniel Hardisky who found out that -- π = 3 + 2/3 ( 1/C[4,3] -1/C[6,3] + 1/C[8,3] + ........] (infinite series) Proof- π = 3 + 4/(2*3*4) -4/(4*5*6) + 4/(6*7*8) - ............. (each of the denominators belong to Nilakantha's series and is the area of a Pythogorean triangle) or π = 3 + 4/6([1*2*3/2*3*4] - [1*2*3/4*5*6] + [1*2*3/6*7*8] +..........) or π = 3 + 2/3 ( 1/C[4,3] -1/C[6,3] + 1/C[8,3] + ........] * identities between Binomial co-efficients are known as " The Star of David theorems" * Harlan brothers recently discovered the fundamental constant e hidden in the Pascal's triangle by analyzing the products in stead of sums of all elements in a row. * Pascal's triangle shows up in many equations and algorithms ranging from simple algebra to finite calculus. * Newton interpolated the binomial coefficients in a recursive manner and discovered the following series-- 1/0! - (1-x)/1! +1*(1-x)*(2-x) /2! - 1*(1-x)*(2-x) *(3-x) /3! +.............. *Leibniz
Series given by π/4 = Σn=0
(-1) * In the same
way, Nilakantha Series : π/8 = 1/ (2 π/8 = 1/ (2 =1/6 ( C(2,1)/C(3,3) + C(6,1)/C(7,3) +C(10,1)/C(11,3) + ..... ) or π = 4/3 ( C(2,1)/C(3,3) + C(6,1)/C(7,3) +C(10,1)/C(11,3) + ..... ) |