The two terms can be immediately summed, since they are fractions with the same denominator:
[tex]\dfrac{6}{n}+\dfrac{2}{n}=\dfrac{6+2}{n}=\dfrac{8}{n}[/tex]
In order to add 1, we must think of it as [tex]\frac{n}{n}[/tex]:
[tex]\dfrac{8}{n}+1=\dfrac{8}{n}+\dfrac{n}{n}=\dfrac{8+n}{n}[/tex]
Answer:
Answer above me is wrong, it's D: 8n+6/n(n+1)
Step-by-step explanation:
For edge.
