What is ∑ K^2(28cK) where K=0 to 28 and 28cK is the number of ways of choosing k items from 28 items?

a. 406 * 2^27
b. 306 * 2^26
c. 28 * 2^27
d. 56 * 2^27

• we know,
  (1+x)^n=nc0+nc1.x^1+nc2.x^2+nc3.x^3+...+ncn.x^n
  
  differentiate both sides w.r.t x
  n(1+x)^n-1=nc1+2.nc2.x+3.nc3.x^2+...+n.ncn.x^n-1
  
  now multiply both sides by x
  x.n(1+x)^n-1=nc1.x+2.nc2.x^2+3.nc3.x^3+...+n.ncn.x^n
  
  again, differentiate both sides w.r.t x
  x.n(n-1)(1+x)^n-2 + n(1+x)^n-1 = 1^2.nc1+2^2.nc2.x+3^2.nc3.x^3+...+n^2.ncn.x^n-1
  
  n(1+x)^n-2[x(n-1)+(1+x)]= 1^2.nc1+2^2.nc2.x+3^2.nc3.x^3+...+n^2.ncn.x^n-1
  
  => n(1+nx)(1+x)^n-2 = 1^2.nc1+2^2.nc2.x+3^2.nc3.x^3+...+n^2.ncn.x^n-1
  
  putting x=1 we get,
  => n(1+n)*2^n-2 = 1^2*nc1+2^2*nc2+3^2*nc3+...+n^2*ncn
  
  so, ∑ K^2(kCk)[K=0 to K=K] = k(k+1)*2^k-2
  
  put k=28 in the above result,
  
  we get, ∑ K^2(28cK) where K=0 to 28 = 28*29*2^26 = 14*29*2^27 = 406*2^27
  
  ans: 406 * 2^27
• 10 years agoHelpfull: Yes(41) No(0)
• = n(n+1)/2
  =28*29/2
  =406
  ans is-> a>406*2^27
• 10 years agoHelpfull: Yes(7) No(6)
• formula = n(n+1)2^n-2 this is formula i dont know its explaination
  
• 10 years agoHelpfull: Yes(5) No(5)
• guyz plz find more puzzels on this web link :http://adityasrmca.blogspot.in/
• 10 years agoHelpfull: Yes(1) No(7)