ΣnCk = 1 (k=0)
ΣnCk = 1 + n (k=0 to k=1)
ΣnCk = 1 + n(n+1)/2 (k=0 to k=2)
ΣnCk = 1 + n(n+1)/2 +( n(n-1)(n-2)/2) - Σn-->3 to n[(n-1)(n-2) + (n-2)(n-3) +...... (n-n)(n-n+1)] (k=0 to k=3)
ΣnCk = 1 + n(n+1)/2 + [n(n-1)(n-2)/2] - Σn-->3 to n[(n-1)(n-2) + (n-2)(n-3) +...... (n-n)(n-n+1)] +1/2 [1*(n-2)(n-3)+2*(n-3)(n-4) +
3*(n-4)(n-5) + .... +(n-3)*(n-(n-2))(n-(n-1)) ] (k=0 to k=4 and (n-3) is the last term co-efficient)
Qk is the ratio of frequency for k and frequency for k-1 i.e Qk =((n+1-k)/k)*(x/y) =qk (x/y)
where qk = (n+1-k)/k
Property of qk and Qk (excluding behavior at 0≤k<1)
*Qk monotonically decreases as k increases, n remaining constant. Qk (maximum) = n*(x/y) for k=1
Qk (minimum) =(1/n)*(x/y) for k=n
*qk monotonically decreases as k increases, n remaining constant. qk (maximum) = n for k=1 (qk maximum value being ≥1)
qk (minimum) =(1/n) for k=n (qk minimum value being <1)
qk =1 when k=(n+1)/2 i.e. when n is an odd number since k has to be an integer.
Qk =1 when k=(n+1)/(z+1) where z = (y/x) and k has to be an integer.
* If we define a function f (x) such that Y = f (x) = (n+1-x)/x (do not confuse Y with small y) , then
Y =(n+1)/x - 1. Define another function z =f (Y) = Y + 1 , then z =f1(x) = (n+1)/x. Since n+1 is a constant say A, z = A/x or zx=A
This is the Equation of a Equilateral or Rectangular Hyperbola with asymptotes taken as the co-ordinate axis and is of the form xy=a2/2 = constant. The coordinates of the vertices are on the bisector of the first and third quadrant and the first and second coordinate coincide, that is to say, x = z. Also, Point A belongs to the curve of the hyperbola.
Vertices lie in 1st,3rd quadrant and are (√A,√A) and (-√A,-√A)
Length of each Semi-Axis (here a=b) a, b is √(2A). Length of Semi -axis c is √(a2+b2) =√(a2+a2) = √(2a2) =√(2.2A) =2√A
Focus is (√(2A),√(2A) ) , (-√(2A),-√(2A) ). Eccentricity=c/a= √2 (Eccentricity of hyperbola is always greater than 1)