what is summation of [3^k(28Ck)] (summation range 28 to k=0) where 28Ck is the number of ways of choosing k items from 28 items??

• Ans is 4^28.
  There is formula summatio X^P(NcP) where p=0,1,2,3,4.......n.=(X+1)^n.
  In this question X=3, N=28. So ans is 4^28.
• 9 years agoHelpfull: Yes(24) No(3)
• The expansion of (1+x)^n=nC0 x^0+nC1 x^1+nC2 x^2+......
  putting the value of x=3 and n=28 we get
  (1+3)^28=28C0 3^0+28C1 3^1+28C2 3^2+28C3 3^3.......
  which is the given summation we need to find hence,
  ans= 4^28 or 2^56
• 9 years agoHelpfull: Yes(11) No(0)
• 4^28
  we knw
  nc0+nc1x+nc2x^2+.......=(1+x)^28
• 9 years agoHelpfull: Yes(1) No(0)
• Answer is : 256
  
  We know that C0+3C1+32C2+.....+3nCn=4n
  Substitute n = 28
  We get ΣK=0283K(28KC) = 428= 256
  
• 9 years agoHelpfull: Yes(0) No(0)