Combinatorics

Question

Somebody prove this identity :

$\frac{\binom{n-1}{0}}{1} + \frac{\binom{n-1}{1}}{2} + \frac{\binom{n-1}{2}}{3} + .... + \frac{\binom{n-1}{n-1}}{n} = \frac{2^{n}-1}{n}$

Comments

see this,

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sorry, i forget to crop the image...irrelevant data is also present in it..

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@nitish Thank you. I was trying to solve it by expanding (x + y)^n-1 and then integrating wrt y

Answer 1

One nice approach which I want to mention in this regard :

We know : 

(1 + x)^n-1     =    ^n-1C₀ . (1)   +    ^n-1C₁ (1) . x + ^n-1C₂ (1) . x²  ....................  + ^n-1C_n-1 . x^n-1             --------(1)

 Now integrating both sides w.r.t. x , we have : 

             (1 + x)^n-1+1 / (n-1+1)    =  [  ^n-1C₀ . x  ]  +   [ ^n-1C₁  . x² / 2 ]  ...............   + [  ^n-1C_n-1 . xⁿ / n ]  + C [ where C is constant of integration as it is an indefinite integral ]                                                                                      -----------(2)

To find the value of C , we substitute x  =  0 in the above equation so by that we obtain C = 1 / n ....

Now substituting x = 1 both sides we have :

             2ⁿ /  n                         =    ^n-1C₀    +  ( ^n-1C₁ / 2 )   +  ....................   +    ^n-1C_n-1 / n    +  C

                                               =    ^n-1C₀    +  ( ^n-1C₁ / 2 )   +  ....................   +    ^n-1C_n-1 / n    +  1/n

==>     (2ⁿ - 1) / n                     =    ^n-1C₀    +  ( ^n-1C₁ / 2 )   +  ....................   +    ^n-1C_n-1 / n

Hence the answer should be (2ⁿ - 1) / n .

Comments

Yes thank you. I had missed the 'C'. stupid mistake -_-

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I was also wondering the same that how I was getting 2ⁿ / n..But then I recalled that in indefinite integral we need to mention a constant of integration..

Here the constant 'C' shows its significance which it does not show normally..:)