SOLUTION: nc1+2nc2+3nc3+...+nncn

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Question 1080977: nc1+2nc2+3nc3+...+nncn
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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nc1+2nc2+3nc3+...+nncn
~~~~~~~~~~~~~~~~~ 

The sum is 

Sum = C%5Bn%5D%5E1 + 2%2AC%5Bn%5D%5E2 + 3%2AC%5Bn%5D%5E3 + . . . + n%2AC%5Bn%5D%5En.


The common term is

k%2AC%5Bn%5D%5Ek =  =  =  = n%2AC%5Bn-1%5D%5E%28n-1%29.

Therefore,

Sum = C%5Bn%5D%5E1 + 2%2AC%5Bn%5D%5E2 + 3%2AC%5Bn%5D%5E3 + . . . + n%2AC%5Bn%5D%5En = n%2AC%5Bn-1%5D%5E0 + n%2AC%5Bn-1%5D%5E1 + n%2AC%5Bn-1%5D%5E2 + . . . n%2AC%5Bn-1%5D%5E%28n-1%29 = .

It is widely known (*) that 

C%5Bn-1%5D%5E0+%2B+C%5Bn-1%5D%5E1+%2B+C%5Bn-1%5D%5E2+%2B+ellipsis+%2B+C%5Bn-1%5D%5E%28n-1%29 = 2%5E%28n-1%29.   (*)

Hence,

Sum = n%2A2%5E%28n-1%29.


Regarding the statement (*) see the lesson
    Remarkable identities for Binomial Coefficients 
in this site.

Solved.