1.

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

Answer»

`(-1)^(n)*"^(3N-1)C_(n-1)`
`(-1)^(n)*"^(3n-1)C_(n)`
`(-1)n*"^(3n-1)C_(n+1)`
None of these

Solution :`(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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