Answer
Clearly, from the above proof, it is evident that the last termin the series of ik is (1/k+1) * nk+1, whichis Θ(nk+1).
Let san- .Then the relevant identity, derived in the same way from the binomial expansion, is k-1 a +1 a + 1 0+1 Sa-1,n This recursive identity gives a formula for sa, in
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