In this following text excerpted from Modern Algebra by Seth Warner, the author showed how a finite linear combination of a sequence of vectors imply a linear combination of the set containing the members of the sequence of the vectors:
If $(a_k)_{1~\leq~ k~\leq ~n}$ is a sequence of elements of a $K$-module $E$, then an element $b$ is a linear combination of the sequence $(a_k)_{1~\leq~ k~\leq ~n}$ if and only if $b$ is a linear combination of $\{a_1, \ldots, a_n\}$. Indeed, every linear combination of $(a_k)_{1~\leq~ k~\leq ~n}$ is clearly a linear combination of $\{a_1, \ldots, a_n\}$. Conversely, let $b$ be a linear combination of $\{a_1, \ldots, a_n\}$. Then there exist a sequence $(c_j)_{1~\leq~ j~\leq ~m}$ of elements of $\{a_1,\ldots, a_n\}$ and a sequence $(\mu_j)_{1~\leq~ j~\leq ~m}$ of scalars such that $b~=~ \displaystyle \sum_{j~=~1}^m \mu_jc_j$. For each $k\in [1,n]$, if $a_k\in \{c_1, \ldots, c_m\}$ and if $a_i \ne a_k$ for all indices $i$ such that $1\leq i\lt k$ , let $\lambda_k$ be the sum of all the scalars $\mu_j$ such that $c_j=a_k$ , but if $a_k\notin\{c_1, \ldots, c_m\}$ or $a_i=a_k$ for some index $i$ such that $1\leq i\lt k$, let $\lambda_k=0$. Then clearly $$b~=~ \sum_{j~=~1}^m \mu_jc_j~=~\sum_{k~=~1}^n \lambda_ka_k$$ ...
However, I couldn't comprehend the bold statement above in the proof.
Why did the author define $\lambda_k$ by that way mentioned above? By taking the sum of the scalars $\mu_j: ~c_j = a_k$, does he mean there are more than one $c_i$s which are identical to each other and equal to $a_k$?