binomial theorem

Question

n-1

∑           ⁿc_r/(ⁿc_r+ⁿc_r+1) equals?

r=0

Answer 1

From the formula for the binomial coefficients :  n!/[r! (n-r)! ] ,

The sum can be rearranged as

 

  Σ^n-1_r=0   1/[1+ (n-r)/(r+1) ]        and then

Σ^n-1 _r=0   (r+1)/(1+n)    ,  the factor of 1/(1+n) comes outside the sum leaving sums easy to evaluate.

 

[1/(1+n)] [n(n-1) n/2 +n]   =   n/2

 

Answer 2

Minakshi:

 

Divide the summand in the numerator and the denominator by ⁿC_r. In the denominator you get 1 plus the following term, which you simplify.

 

[ⁿC_r+1 / ⁿC_r] = [(n!)/(n-r-1)!(r+1)!)] / [(n!)/(n - r)!(r)!] = (n-r)/(r+1)

 

1+[ⁿC_r+1 / ⁿC_r] = (n+1)/(r+1)

 

and

 

1 / {1+[ ⁿC_r+1 /ⁿC_r]} = (r+1)/(n+1)

 

Therefore, your required sum equals [1/(n+1)] times the sum of the first n positive integers, 1, 2, ..., n. This latter sum is n(n+1)/2.  (You can prove this by induction.)  Hence the required sum equals n/2.

 

Post a comment please.

 

With blessings,

 

Dr. G.

 

 

(n+1) times the sum of the reciprocals of the first n positive integers. The sum of the reciprocals of the first n positive integers is called the Harmonic Number H_n.  (Google "harmonic number" to read more).  So your required sum can be written as (n+1)H_n.