To expand the given summation, we proceed as follows:
[tex]\begin{gathered} \text{Given:} \\ \sum ^6_{n\mathop=1}(n-1) \\ \Rightarrow\text{ }\sum ^6_{n\mathop{=}1}(n)-\text{ }\sum ^6_{n\mathop{=}1}(1) \\ \text{Now:} \\ \sum ^6_{n\mathop{=}1}(n)\text{ is the sum of the first six natural numbers (1,2,3,4,5,6)} \\ \text{And:} \\ \sum ^6_{n\mathop{=}1}(1)\text{ is simply (6}\times1)--That\text{ is, the number 1 added to itself six times } \\ \text{Therefore, we have:} \\ \Rightarrow\text{ }\sum ^6_{n\mathop{=}1}(n)-\text{ }\sum ^6_{n\mathop{=}1}(1) \\ \Rightarrow(1+2+3+4+5+6)-(1+1+1+1+1+1) \\ \Rightarrow(1+2+3+4+5+6)-(6) \\ \Rightarrow(1+2+3+4+5) \\ \end{gathered}[/tex]Therefore:
[tex]\sum ^6_{n\mathop{=}1}(n-1)\text{ = 1+2+3+4+5}[/tex]So, the correct option is option C
This is because the sum: 0+1+2+3+4+5 gives the same value as the sum: 1+2+3+4+5