Some sequences are defined by a recursive formula- that is, a formula that defines each term of the sequence in terms of one or more of the preceding terms. For example, if a n is defined by a 1 = 1 and a n = 2 a n − 1 + 1 for n ≥ 2 Then a 2 = 2 a 1 + 1 = 2 ⋅ 1 + 1 = 3 a 3 = 2 a 2 + 1 = 2 ⋅ 3 + 1 = 7 a 4 = 2 a 3 + 1 = 2 ⋅ 7 + 1 = 15 and so on. In Problem 63 - 66 , write the first five terms of each sequence. a 1 = 2 and a n = 3 a n − 1 + 2 for n ≥ 2
Some sequences are defined by a recursive formula- that is, a formula that defines each term of the sequence in terms of one or more of the preceding terms. For example, if a n is defined by a 1 = 1 and a n = 2 a n − 1 + 1 for n ≥ 2 Then a 2 = 2 a 1 + 1 = 2 ⋅ 1 + 1 = 3 a 3 = 2 a 2 + 1 = 2 ⋅ 3 + 1 = 7 a 4 = 2 a 3 + 1 = 2 ⋅ 7 + 1 = 15 and so on. In Problem 63 - 66 , write the first five terms of each sequence. a 1 = 2 and a n = 3 a n − 1 + 2 for n ≥ 2
Solution Summary: The author calculates the first five terms of the sequence a_1=2, and the second term by using a recursive formula.
Some sequences are defined by a recursive formula- that is, a formula that defines each term of the sequence in terms of one or more of the preceding terms. For example, if
a
n
is defined by
a
1
=
1
and
a
n
=
2
a
n
−
1
+
1
for
n
≥
2
Then
a
2
=
2
a
1
+
1
=
2
⋅
1
+
1
=
3
a
3
=
2
a
2
+
1
=
2
⋅
3
+
1
=
7
a
4
=
2
a
3
+
1
=
2
⋅
7
+
1
=
15
and so on. In Problem
63
-
66
, write the first five terms of each sequence.
Determine the number of automorphisms of the following graph. Explain why your
answer is correct.
Find the bisector of the angle <ABC in the Poincaré plane, where A=(0,5), B=(0,3) and C=(2,\sqrt{21})
The masses measured on a population of 100 animals were grouped in the
following table, after being recorded to the nearest gram
Mass
89 90-109 110-129 130-149 150-169 170-189 > 190
Frequency 3
7 34
43
10
2
1
You are given that the sample mean of the data is 131.5 and the sample
standard deviation is 20.0. Test the hypothesis that the distribution of masses
follows a normal distribution at the 5% significance level.
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