Answer:
[tex]\large\boxed{-11,325}[/tex]
Step-by-step explanation:
First simplify:
[tex]-1-(n-1)=-1-n-(-1)=-1-n+1=-n[/tex]
Therefore we have:
[tex]\sum\limits_{n=1}^{150}[-1-(n-1)]=\sum\limits_{n=1}^{150}(-n)=(-1)+(-2)+(-3)+...+(-150)\\\\-1,\ -2,\ -3,\ -4,\ ...,\ -150-\text{it's the arithmetic sequence}\\\text{with the common difference d = -1.}\\\\\text{The formula of a sum of terms of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=150,\ a_1=-1,\ a_n=-150:\\\\S_{150}=\dfrac{-1+(-150)}{2}\cdot150=(-151)(75)=-11,325[/tex]
