Answer:
1680
Step-by-step explanation:
[tex]\sum\limits_{n=1}^{42}2n-3\\\\a_n=2n-3\\\\a_{n+1}=2(n+1)-3=2n+2-3=2n-1\\\\a_{n+1}-a_n=(2n-1)-(2n-3)=2n-1-2n+3=2=constant\\\\\text{Therefore it's an arithmetic sequence with first term}\\\\a_1=2(1)-3=2-3=-1\\\\\text{and common difference}\\\\d=2[/tex]
[tex]\text{The formula of a sum of terms of an aithmetic sequence:}\\\\S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n\\\\\text{Substitute:}\\\\n=42,\ a_1=-1,\ d=2:\\\\S_{42}=\dfrac{2(-1)+(42-1)(2)}{2}\cdot42=[-2+(41)(2)](21)=(-2+82)(21)\\\\=(80)(21)=1680[/tex]
