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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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