Question

can someone explain to me how to do this : an = ½ an-1 , where a1 = 20

Answer 1

\(a_{n} = f(a_{n-1})\) MEANS the following sequence...

\(a_{2} = f(a_{1})\)
\(a_{3} = f(a_{2})\)
\(a_{4} = f(a_{3})\)
\(a_{5} = f(a_{4})\)
\(a_{6} = f(a_{5})\)
...

You are given \(a_{1}\). Apply the formula (multiply by 1/2) to obtain \(a_{2}\)
You now have \(a_{2}\). Apply the formula (multiply by 1/2) to obtain \(a_{3}\)
and so forth...

Answer 2

soo all i have to do is multiply 20 by 1/2 ?

Answer 3

...and that will be \(a_{2}\). Do it again to find \(a_{3}\).

Answer 4

Oftentimes, the formula is more complicated.

Answer 5

wait a sec

Answer 6

okay so I got 10 for a2 is that right?

Answer 7

\(a_{1} = 20\)

\(a_{2} = (1/2)a_{1}\)

\(a_{2} = (1/2)\cdot 20 = 10\)

\(a_{2} = 10\)

I think you have it. Why do you doubt?

Answer 8

okay, im still confused is that the answer of -->an = ½ an-1<--

Answer 9

You missed an important part of the discussion. Repeated here for your convenience.

"MEANS the following sequence..."

That notation suggests a sequence of events. Another, perhaps more clear, way to write it might be,

\(a_{NextInSequence} = (1/2)⋅(a_{PreviousInSequence})\)

You need only a place to start, such as a1=20, in order to define the entire SEQUENCE. To generate the ENTIRE sequence, you must apply the formula infinitely many times.

Answer 10

\(\bf \large{a_{\color{red}{ n}}=\cfrac{1}{2}a_{{\color{red}{ n}}-1}\qquad \qquad a_1=20\\ \quad \\

a_{\color{red}{ 2}}=\cfrac{1}{2}a_{{\color{red}{ 2}}-1}\implies a_{\color{red}{ 2}}=\cfrac{1}{2}a_{{\color{red}{ 1}}}\qquad a_{\color{red}{ 3}}=\cfrac{1}{2}a_{{\color{red}{ 3}}-1}\implies \cfrac{1}{2}a_{{\color{red}{ 2}}}\\ \quad \\

a_{\color{red}{ 7}}=\cfrac{1}{2}a_{{\color{red}{ 7}}-1}\implies \cfrac{1}{2}a_{{\color{red}{ 6}}}\qquad
a_{\color{red}{ 9}}=\cfrac{1}{2}a_{{\color{red}{ 9}}-1}\implies \cfrac{1}{2}a_{{\color{red}{ 8}}}\quad ...}\)

Answer 11

ohh okay i get it now thank you yall