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A scientist observes a culture of bacteria. Initially (at day 0), there are \(u_0=5\) bacteria. Each day, the number of bacteria doubles. Let \(u_n\) be the number of bacteria at day \(n\).
Part A: Define the Sequence Recursively
The initial term is \(u_0 =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
The recursive rule is \(u_{n+1} =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
\(\times u_n \).
Part B: Calculate the Terms for the First Five Days
\(u_1 =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
bacteria
\(u_2 =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
bacteria
\(u_3 =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
bacteria
\(u_4 =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
bacteria
\(u_5 =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
bacteria
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