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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??
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- 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. - 10 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 - 10 years agoHelpfull: Yes(11) No(0)
- 4^28
we knw
nc0+nc1x+nc2x^2+.......=(1+x)^28 - 10 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
- 10 years agoHelpfull: Yes(0) No(0)
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