What is the sum of 28 terms in the following expressions :
(3^k)*(28Ck),
where (28Ck) is the number of ways of choosing k items from 28 items?
a) 2^56
b) 3*(2^27)
c) 3^29
d) 3*(4^27)

• Tanu its for u...
  we know that (1+x)^n=nc0+nc1.x^1+nc2.x^2+....+ncn.x^n
  here x=3 and n=28
  hence, 28c0+28C1.3^1+28C2.3^2+.....28C28.3^28 can be written as(1+3)^28
  therefore,(4)^28=(2)^56
  so... 2^56
• 10 years agoHelpfull: Yes(20) No(1)
• 28C1.3^1+28C2.3^2+.....28C28.3^28
  this can be written as (1+3)^28 - 1
  so..4^28 -1
  hence,3.4^27 is the answer
• 10 years agoHelpfull: Yes(11) No(10)
• plz tel me the correct ans with xplanatn
• 10 years agoHelpfull: Yes(6) No(0)
• (1+n)^28= sum of ((n^k)* (28Ck))
  Now put n=3
  So answer= 4^23=2^46
  
• 10 years agoHelpfull: Yes(5) No(6)