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The inductive hypothesis is the temporary assumption made during the inductive step that the proposition $P(k)$ is true for an arbitrary integer k. It is used as a premise to derive the truth of $P(k+1)$.

Conditions: Used within the proof of the inductive step.; Assumes $P(k)$ is true for a fixed arbitrary k.; Required to prove $P(k+1)$.

The inductive hypothesis is the temporary assumption made during the inductive step that the proposition $P(k)$ is true for an arbitrary integer k. It is used as a premise to derive the truth of $P(k+1)$.

Conditions: Used within the proof of the inductive step.; Assumes $P(k)$ is true for a fixed arbitrary k.; Required to prove $P(k+1)$.

The proof uses mathematical induction. First, the base case is verified by checking that $a_1=1 \ge \frac{1}{2}$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n \ge 1$.; The initial value is $a_1=1$.; The goal is to prove $a_n \ge \frac{1}{2}$ for all $n \ge 1$.

The proof uses mathematical induction. First, the base case is verified by checking that $a_1=1 \ge \frac{1}{2}$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n \ge 1$.; The initial value is $a_1=1$.; The goal is to prove $a_n \ge \frac{1}{2}$ for all $n \ge 1$.

The two conditions are: 1) The base case, where the propositional statement $P(1)$ is verified as true for the initial value $n=1$. 2) The inductive step, where it is proven that if the statement is true for an arbitrary integer $k (P(k))$, then it must also be true for $k+1 (P(k+1))$.

Conditions: The statement is a proposition $P(n)$ indexed by natural numbers.; The base case is typically checked at $n=1$.; The inductive step assumes $P(k)$ is true to prove $P(k+1)$.

The two conditions are: 1) The base case, where the propositional statement $P(1)$ is verified as true for the initial value $n=1$. 2) The inductive step, where it is proven that if the statement is true for an arbitrary integer $k (P(k))$, then it must also be true for $k+1 (P(k+1))$.

Conditions: The statement is a proposition $P(n)$ indexed by natural numbers.; The base case is typically checked at $n=1$.; The inductive step assumes $P(k)$ is true to prove $P(k+1)$.

Assume $P(k)$ is true: $2 + 4$ + ... + $2k = k(k+1)$. To prove $P(k+1)$, write the LHS as the sum up to 2k plus the next term $2(k+1)$.

Conditions: Inductive hypothesis: $P(k)$ is true.; Goal: Prove $P(k+1)$ is true.; Algebraic manipulation involves factoring common terms.

Assume $P(k)$ is true: $2 + 4$ + ... + $2k = k(k+1)$. To prove $P(k+1)$, write the LHS as the sum up to 2k plus the next term $2(k+1)$.

Conditions: Inductive hypothesis: $P(k)$ is true.; Goal: Prove $P(k+1)$ is true.; Algebraic manipulation involves factoring common terms.