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Sum of the series sum_(k=0)^(n)""^(n)C_(k)(-1)^(k)1/(a_(k)) where a_(k)=sum_(i=0)^(k)""^(k)C_(i)b_(i) where b_(i)=sum_(j=0)^(i)""^(i)C_(j)((-2)/3)^(j) is |
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Answer» `1/(2^(N))` |
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