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The Principle of Mathematical Induction states that if a proposition $P(n)$ is true for the base case (usually $n=1$) and if the truth of $P(k)$ implies the truth of $P(k+1)$ for any arbitrary integer k, then $P(n)$ is true for all natural numbers n.

Conditions: $P(1)$ is true.; For any k, $P(k)$ implies $P(k+1)$.; n is a natural number.

The Principle of Mathematical Induction states that if a proposition $P(n)$ is true for the base case (usually $n=1$) and if the truth of $P(k)$ implies the truth of $P(k+1)$ for any arbitrary integer k, then $P(n)$ is true for all natural numbers n.

Conditions: $P(1)$ is true.; For any k, $P(k)$ implies $P(k+1)$.; n is a natural number.

This rewrite is possible due to the distributive property of multiplication over addition, specifically factoring out the common term ($k+1$). Both terms in the sum $k(k+1)$ and $2(k+1)$ share the factor ($k+1$).

Conditions: The expression is a sum of two terms: $k(k+1)$ and $2(k+1)$.; ($k+1$) is a common factor in both terms.

This rewrite is possible due to the distributive property of multiplication over addition, specifically factoring out the common term ($k+1$). Both terms in the sum $k(k+1)$ and $2(k+1)$ share the factor ($k+1$).

Conditions: The expression is a sum of two terms: $k(k+1)$ and $2(k+1)$.; ($k+1$) is a common factor in both terms.

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 $P(k+1)$.