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Write the first five terms of the sequence with the given nth term. an = 3n − 1 n!.

Sagot :

The first five terms of the sequence with the given nth term are 1, 3/2, 3/2, 9/8, 27/40.

In this question,

The nth term of the sequence is [tex]a_n=\frac{3^{(n-1)} }{n!}[/tex]

Now substitute the values of n = 1,2,3,4,5

For n = 1,

⇒ [tex]a_1=\frac{3^{(1-1)} }{1!}[/tex]

⇒ [tex]a_1=\frac{3^{0} }{1}[/tex]

⇒ a₁ = 1

For n = 2,

⇒ [tex]a_2=\frac{3^{(2-1)} }{2!}[/tex]

⇒ [tex]a_2=\frac{3^{(1)} }{(1)(2)}[/tex]

⇒ [tex]a_2=\frac{3}{2}[/tex]

For n = 3,

⇒ [tex]a_3=\frac{3^{(3-1)} }{3!}[/tex]

⇒ [tex]a_3=\frac{3^{(2)} }{(1)(2)(3)}[/tex]

⇒ [tex]a_3=\frac{9}{6}[/tex]

⇒ [tex]a_3=\frac{3}{2}[/tex]

For n = 4,

⇒ [tex]a_4=\frac{3^{(4-1)} }{4!}[/tex]

⇒ [tex]a_4=\frac{3^{(3)} }{(1)(2)(3)(4)}[/tex]

⇒ [tex]a_4=\frac{27}{24}[/tex]

⇒ [tex]a_4=\frac{9}{8}[/tex]

For n = 5,

⇒ [tex]a_5=\frac{3^{(5-1)} }{5!}[/tex]

⇒ [tex]a_5=\frac{3^{(4)} }{(1)(2)(3)(4)(5)}[/tex]

⇒ [tex]a_5=\frac{81}{120}[/tex]

⇒ [tex]a_5=\frac{27}{40}[/tex]

Hence we can conclude that the first five terms of the sequence with the given nth term are 1, 3/2, 3/2, 9/8, 27/40.

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