Respuesta :
Answer:
Option C) [tex]a_{n}=m-b+m(n-1)[/tex] for [tex]n={\{1,2,3,..}\}[/tex]
is the answer
Step-by-step explanation:
Given infinite sequence is [tex]{\{m+b,2m+b,3m+b,...}\}[/tex]
[tex]a_{n}=m-b+m(n-1)[/tex] for [tex]n={\{1,2,3,..}\}[/tex] is not the correct way to define the given infinite sequence
Put n=1 in [tex]a_{n}=m-b+m(n-1)[/tex]
[tex]a_{1}=m-b+m(1-1)[/tex]
[tex]=m-b[/tex]
[tex]a_{1}=m-b[/tex]
Put n=2 in [tex]a_{n}=m-b+m(n-1)[/tex]
[tex]a_{2}=m-b+m(2-1)[/tex]
[tex]=m-b+m[/tex]
[tex]a_{2}=2m-b[/tex]
Put n=3 in [tex]a_{n}=m-b+m(n-1)[/tex]
[tex]a_{3}=m-b+m(3-1)[/tex]
[tex]=m-b+2m[/tex]
[tex]a_{3}=3m-b[/tex]
Therefore the infinite sequence [tex]{\{m-b.2m-b,3m-b,...}\}[/tex] for [tex]a_{n}=m-b+m(n-1)[/tex] for [tex]n={\{1,2,3,..}\}[/tex] is not same as the given infinite sequence [tex]{\{m+b,2m+b,3m+b,...}\}[/tex]
Therefore option C) [tex]a_{n}=m-b+m(n-1)[/tex] for [tex]n={\{1,2,3,..}\}[/tex]
is the answer
Answer:
C if your doing prep
Step-by-step explanation:
math