How do you write the first five terms of the sequence defined recursively #a_1=32, a_(k+1)=1/2a_k#?

Question

Anna Allen · Accepted Answer

#a_1 = 32, a_2 = 16, a_3 = 8, a_4 = 4, a_5 = 2# When #k=1, a_(k+1) = a_2 = 1/2 a_1 = 1/2(32) = 16# When #k=2, a_(k+1) = a_3 = 1/2 a_2 = 1/2(16) = 8# When #k=3, a_(k+1) = a_4 = 1/2 a_3 = 1/2(8) = 4# When #k=4, a_(k+1) = a_5 = 1/2 a_4 = 1/2(4) = 2# So: #a_1 = 32, a_2 = 16, a_3 = 8, a_4 = 4, a_5 = 2#

Theodore Amidon · Answer

undefined To find the first five terms of the sequence defined recursively by $a_1 = 32$ and $a_{k+1} = \frac{1}{2}a_k$, we can use the recursive definition of the sequence to generate each subsequent term.

1. $a_1 = 32$
2. $a_2 = \frac{1}{2}a_1 = \frac{1}{2}(32) = 16$
3. $a_3 = \frac{1}{2}a_2 = \frac{1}{2}(16) = 8$
4. $a_4 = \frac{1}{2}a_3 = \frac{1}{2}(8) = 4$
5. $a_5 = \frac{1}{2}a_4 = \frac{1}{2}(4) = 2$

So, the first five terms of the sequence are $32$, $16$, $8$, $4$, and $2$.