1 Question Algebra 2

- anonymous

1 Question Algebra 2

- chestercat

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- anonymous

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- UsukiDoll

Recursion is the process of choosing a starting term and repeatedly applying the same process to each term to arrive at the following term. Recursion requires that you know the value of the term immediately before the term you are trying to find.

- Jack1

yay, cheers @KimberlyAlice and @UsukiDoll ;P

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## More answers

- anonymous

Okay, so how would I find the equation for this problem?
Thanks @Jack1 :P

- UsukiDoll

so examples of recursion formulas are
\[a_1 = 4, a_n = 2a_{n-1}\]
or
\[a_1=4 , a_{n+1} = 2a_n\]

- UsukiDoll

so what is the \[a_1 \] in this problem?

- UsukiDoll

we need to know what a_1 is or we are stuck

- anonymous

a1 would be the recursive number?

- UsukiDoll

well we need to figure out the next term it's like a chain.. but we need it to start at \[a_n \] where n = 1

- UsukiDoll

what's the starting number? that's a_n when n =1

- anonymous

7

- UsukiDoll

ok.. so \[\large a_1=7\]

- anonymous

That limits the choices to B and D

- UsukiDoll

so now we need to figure out ... we need to get to 4 somehow
but what recursion formula will allow us to do that

- anonymous

B

- anonymous

No wait...we would get 3

- UsukiDoll

yeah that's what I mean.. we need to get to 4's land

- UsukiDoll

we know the difference in each number is 3

- anonymous

Yes

- anonymous

So it would be B?

- UsukiDoll

let me think...

- UsukiDoll

alright our starting point is \[a_1 = 7 \] so now we need a formula to get to 4

- UsukiDoll

OH how I wish I can do this
\[a_n=a_{n-1}-3 \] where n = 2
\[a_2=a_{2-1}-3 \]
\[a_2=a_{1}-3 \]
since \[a_1 = 7\]
\[a_2=7-3 \]
\[a_2=4 \]

- UsukiDoll

ok I think the multiple choices are flawed. Just look what I've done.. I got a_2 to appear as 4

- UsukiDoll

hey! There's a typo in your choices. You want the last choice with a - sign instead

- anonymous

oh no...

- UsukiDoll

And I'll prove it
so now we have \[a_1 = 7, a_2 =4\]
so now our next recursive formula
\[a_n=a_{n-1}-3 \]
let n = 3
\[a_3=a_{3-1}-3\]
\[a_3=a_{2}-3\]
since \[a_2=4 \]
\[a_3=4-3\]
\[a_3=1\]

- UsukiDoll

see the pattern ... Now I have 7 4 1

- UsukiDoll

so I have to do this 3 more times... for -2 -5 and ?!

- anonymous

I'm confused...which answer choice is this?

- UsukiDoll

or maybe not..
\[a_4 = -2, a_5 = -5\]

- UsukiDoll

there's a typo in the choice we need

- anonymous

But it would be D?

- UsukiDoll

but.. .we can use the formula I have to get our next number.. we know that a_1 = 7 we need a_6

- UsukiDoll

\[{a_6}=a_{6-1}-3\]
\[{a_6}=a_{5}-3\]
since \[a_5= -5\]
\[{a_6}=-5-3\]
\[{a_6}=-8\]

- UsukiDoll

It's the last choice with that typo that shouldn't be there.!!!!

- UsukiDoll

because should that stand... a_1 would've been 10 if it was subtraction it would've been 4

- Jack1

why not \(\Large a_n = 10 -3(n)\)?

- Jack1

@UsukiDoll doesnt that fit?

- UsukiDoll

look at the choices we are given

- Jack1

ah, my bad, sorry :/

- UsukiDoll

it's the last choice with the typo. trust me! We got the beginning which is a_1 is 7
we needed to find a_6 which is -8

- anonymous

Thank you for all your help @Jack1 and @UsukiDoll

- Jack1

np but psh, was all @UsukiDoll , i did nothing

- UsukiDoll

"because should that stand... a_1 would've been 10 if it was subtraction it would've been 4 "
oy I meant a_2 would've been 10 if we had a +
a_1 = 7
a_2 =4
a_3 =1
a_4 =-2
a_5 = -5
a_6 = -8
but the formula is
\[\large a_n=a_{n-1}-3 \]

- anonymous

I understand. thank you!

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