SOLUTION: Proof by induction. Imagine that we are going to prove by induction that:
(1/sqrt(1)) + (1/sqrt(2)) + (1/sqrt(3)) + ... + (1/sqrt(n)) >= sqrt(n), for all n E Z^+
Assume by th
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-> SOLUTION: Proof by induction. Imagine that we are going to prove by induction that:
(1/sqrt(1)) + (1/sqrt(2)) + (1/sqrt(3)) + ... + (1/sqrt(n)) >= sqrt(n), for all n E Z^+
Assume by th
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Question 1176890: Proof by induction. Imagine that we are going to prove by induction that:
(1/sqrt(1)) + (1/sqrt(2)) + (1/sqrt(3)) + ... + (1/sqrt(n)) >= sqrt(n), for all n E Z^+
Assume by the inductive step that:
(1/sqrt(1)) + (1/sqrt(2)) + (1/sqrt(3)) + ... + (1/sqrt(k)) >= sqrt(k), for some k E Z^+
Which of the following is a correct way of ending this proof?
a. (1/sqrt(1)) + (1/sqrt(2)) + ... + (1/sqrt(k)) >= sqrt(k) + (1/sqrt(k+1)) = sqrt(k+1) +1 >= sqrt(k+1)
b. (1/sqrt(1)) + (1/sqrt(2)) + ... + (1/sqrt(k)) >= sqrt(k+1) + (1/sqrt(k+1)) + (1/sqrt(k+1)) >= sqrt(k+1)
c. (1/sqrt(1)) + (1/sqrt(2)) + ... + (1/sqrt(k+1)) >= sqrt(k) + (1/sqrt(k+1)) = (sqrt(k)sqrt(k+1)+1)/sqrt(k+1)) >= (sqrt(k)sqrt(k)+1)/sqrt(k+1)) >= ((k+1)/sqrt(k+1)) = sqrt(k+1)
d. (1/sqrt(1)) + (1/sqrt(2)) + ... + (1/sqrt(k+1)) >= sqrt(k) + (1/sqrt(k)) = ((sqrt(k)sqrt(k+1))/sqrt(k)) >= ((k+1)/(sqrt(k+1))) = sqrt(k+1) Answer by ikleyn(52777) (Show Source):
The proof (c) corresponds to the standard logic of the Mathematical induction method and corresponds
to the logic of the proof the necessary inequality.
