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A recursion formula and the initial term of a sequence are given. Write out the first five terms of the sequence.

[tex]\[
a_1=1, \quad a_{n+1}=\frac{a_n}{n+5}
\][/tex]

A. [tex]\(1, \frac{1}{6}, \frac{6}{7}, \frac{7}{8}, \frac{8}{9}\)[/tex]

B. [tex]\(1, \frac{1}{6}, \frac{1}{42}, \frac{1}{336}, \frac{1}{3,024}\)[/tex]

C. [tex]\(1, \frac{1}{6}, 42, \frac{1}{336}, 3,024\)[/tex]

D. [tex]\(1, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}\)[/tex]



Answer :

GinnyAnswer
Let's analyze the given sequence with the initial term and recursive formula to find the first five terms.

The initial term is given by:
[tex]\[ a_1 = 1 \][/tex]

Now, we use the recursive formula to find the subsequent terms:
[tex]\[ a_{n+1} = \frac{a_n}{n+5} \][/tex]

We start from [tex]\( n = 1 \)[/tex] and proceed step by step:

1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_2 = \frac{a_1}{1+5} = \frac{1}{6} \][/tex]
So, the second term is [tex]\( \frac{1}{6} \)[/tex].

2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_3 = \frac{a_2}{2+5} = \frac{\frac{1}{6}}{7} = \frac{1}{6 \times 7} = \frac{1}{42} \][/tex]
So, the third term is [tex]\( \frac{1}{42} \)[/tex].

3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_4 = \frac{a_3}{3+5} = \frac{\frac{1}{42}}{8} = \frac{1}{42 \times 8} = \frac{1}{336} \][/tex]
So, the fourth term is [tex]\( \frac{1}{336} \)[/tex].

4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ a_5 = \frac{a_4}{4+5} = \frac{\frac{1}{336}}{9} = \frac{1}{336 \times 9} = \frac{1}{3024} \][/tex]
So, the fifth term is [tex]\( \frac{1}{3024} \)[/tex].

Therefore, the first five terms of the sequence are:
[tex]\[ 1, \frac{1}{6}, \frac{1}{42}, \frac{1}{336}, \frac{1}{3024} \][/tex]

Thus, the correct answer is:

B. [tex]\( \boxed{1, \frac{1}{6}, \frac{1}{42}, \frac{1}{336}, \frac{1}{3024}} \)[/tex]

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