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