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Meehan98
 one year ago
Can someone explain to me how to do this, please!
This lesson really confused me.
Identify a possible explicit rule for the nth term of the sequence 1, 1/3, 1/5, 1/7, 1/9, ….
Meehan98
 one year ago
Can someone explain to me how to do this, please! This lesson really confused me. Identify a possible explicit rule for the nth term of the sequence 1, 1/3, 1/5, 1/7, 1/9, ….

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Meehan98
 one year ago
Best ResponseYou've already chosen the best response.0The choices are: \[a _{n}=\frac{ n }{ 2n1 }\] \[a _{n}=\frac{ n }{ 2n+1 }\] \[a _{n}=\frac{ 1 }{ 2n1 }\] \[a _{n}=\frac{ 1 }{ 2n+1 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.. it's asking, on "what pattern are the terms in the sequence show?"

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.0Is the first step to figure out the first differences?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to check what pattern is displaying

Meehan98
 one year ago
Best ResponseYou've already chosen the best response.0Okay, but they aren't constant differences.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1First, think about your sequence like this:\[\large\rm \frac{1}{1},~\frac{1}{3},~\frac{1}{5},~\frac{1}{7},...\]If you rewrite the first number like that, it might help you to see what is going on. They are all `odd` denominators. \(\large\rm 2n1\) and \(\large\rm 2n+1\) are two ways of writing odd numbers. Check out \(\large\rm 2n+1\). If we start counting from n=1, 2(1)+1=3 2(2)+1=5 2(2)+1+7 How bout the other one? \(\large\rm 2n1\). Again if we start counting from n=1, 2(1)1=1 2(2)1=3 2(3)1=5 Hmm these second set of numbers seem to match our denominators, yes?
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