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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 equation $A\vec{x}=\vec{b}$ has no solution because solving it is equivalent to finding weights $x_1, \dots, x_k$ such that $x_1\vec{a}_1 + \dots + x_k\vec{a}_k = \vec{b}$. The column space $C(A)$ is defined as the set of all possible linear combinations of the columns of $A$.

Conditions: $A$ is an $n \times k$ matrix; $\vec{x} \in \mathbb{R}^k$ and $\vec{b} \in \mathbb{R}^n$; $C(A)$ denotes the column space of $A$

The equation $A\vec{x}=\vec{b}$ has no solution because solving it is equivalent to finding weights $x_1, \dots, x_k$ such that $x_1\vec{a}_1 + \dots + x_k\vec{a}_k = \vec{b}$. The column space $C(A)$ is defined as the set of all possible linear combinations of the columns of $A$.

Conditions: $A$ is an $n \times k$ matrix; $\vec{x} \in \mathbb{R}^k$ and $\vec{b} \in \mathbb{R}^n$; $C(A)$ denotes the column space of $A$