Question

Find the nth term of the geometric sequence where:
a1=4 and a_k+1=1/2 a_k. PLEASE

Answer 1

retype the question again

Answer 2

Find the nth term of the geometric sequence where:
a1=4 and a_k+1=1/2 a_k. can u just tell me if I am right? I got : a_n = 4 (1/2)^(n - 1) <---- or in more simplified terms; 2^(n-1)

Answer 3

\[a_{n} = \frac{1}{2}a_{n-1}\]
\[a_{n} = \frac{1}{2}*\frac{1}{2}a_{n-2} = \frac{1}{2^{2}}a_{n-2}\]
continue replacing that a_(n-r) with 0.5*a_(n-r-1) till you reach a_(1)
\[a_{n} = \frac{1}{2^{n-1}}a_{n-(n-1)}\]
\[a_{n} = \frac{1}{2^{n-1}}4 = \frac{1}{2^{n-3}}\]

Answer 4

4(1/2)^(n-1) is correct .. How did you get 2^(n-1) ?

Answer 5

multplied 4 and 1/2

Answer 6

should I just leave it as 4(1/2)^(n-1)?

Answer 7

It isn't (4*1/2)^(n-1) it is (1/2)^(n-1)*4

Answer 8

Ok so is my answer 4(1/2)^(n-1)? @FoolAroundMath

Answer 9

yes. You can simplify it to be 2^(3-n) or (1/2)^(n-3).
4(1/2)^(n-1) is absolutely fine

Answer 10

ok thx can u help me with this @FoolAroundMath : When using math induction to prove that (n3 - n + 3) is divisible by 3 for all natural numbers. First, we will assume that (k3 - k + 3) is divisible by 3. Then, we would need to show that ____ is divisible by 3.

Answer 11

choices k3 - k + 3 + k + 1

(k+1)3 - k + 3

(k+1)3 - (k+1) + 3

k3 - k + 3 + 1

Answer 12

@jim_thompson5910

Answer 13

I think it is (k+1)3 - (k+1) + 3

Answer 14

Proof by Induction:
You assume that it is true for any arbitrary number k.
Then you need to show that it is true for k+1 as well.
So, yes you are correct

Answer 15

You're right, this is because you're testing the k+1 case