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The sum S(n)=sum(k=0)^(n)(-1)^(k)*^(3n)C...

The sum `S_(n)=sum_(k=0)^(n)(-1)^(k)*^(3n)C_(k)`, where `n=1,2,….` is

A

`(-1)^(n)*"^(3n-1)C_(n-1)`

B

`(-1)^(n)*"^(3n-1)C_(n)`

C

`(-1)n*"^(3n-1)C_(n+1)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` `S_(n)=^(3n)C_(0)-^(3n)C_(1)+^(3n)C_(2)+….+(-1)^(n)*^(3n)C_(n)`
But `.^(3n)C_(0)=.^(3n-1)C_(0)`
`-^(3n)C_(1)=-^(3n-1)C_(0)-^(3n-1)C_(1)`
`-^(3n)C_(2)=^(3n-1)C_(1)+^(3n-1)C_(2)`
`-^(3n)C_(3)=-^(3n-1)C_(2)-^(3n-1)C_(3)`
`…………....................................`
`(-1)^(n)*^(3n)C_(n)=(-1)^(n)*^(3n-1)C_(n-1)+(-1)^(n)*^(3n-1)C_(n)`
On adding we get `S_(n)=(-1)^(n)*^(3n-1)C_(n)`
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