Which formula is not equivalent to the other two?

$($a$){\displaystyle \underset{k=6}{\overset{8}{}}}\frac{(-1{)}^{k-5}}{k-5}$

$($b$){\displaystyle \underset{k=0}{\overset{2}{}}}\frac{(-1{)}^k}{k+1}$

$($c$){\displaystyle \underset{k=-5}{\overset{-3}{}}}\frac{(-1{)}^k}{k+6}$

Choose the correct answer below.

${\displaystyle \underset{k=6}{\overset{8}{}}}\frac{(-1{)}^{k-5}}{k-5}$

${\displaystyle \underset{k=0}{\overset{2}{}}}\frac{(-1{)}^k}{k+1}$

${\displaystyle \underset{k=-5}{\overset{-3}{}}}\frac{(-1{)}^k}{k+6}$