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If [tex]$t_1 = 1, t_n = \frac{t_{n-1}}{n} \, (n \geq 2)$[/tex], then [tex]$t_3 = \qquad$[/tex]

1) [tex][tex]$\frac{1}{2}$[/tex][/tex]

2) [tex]$\frac{1}{6}$[/tex]

Sagot :

To determine the value of [tex]\( t_3 \)[/tex] given the recurrence relation [tex]\( t_n = \frac{t_{n-1}}{n} \)[/tex] with the initial term [tex]\( t_1 = 1 \)[/tex], follow these steps:

1. Calculate [tex]\( t_2 \)[/tex]:

Using the recurrence relation for [tex]\( n = 2 \)[/tex],
[tex]\[ t_2 = \frac{t_1}{2} \][/tex]

Substitute [tex]\( t_1 = 1 \)[/tex],
[tex]\[ t_2 = \frac{1}{2} \][/tex]

2. Calculate [tex]\( t_3 \)[/tex]:

Using the recurrence relation for [tex]\( n = 3 \)[/tex],
[tex]\[ t_3 = \frac{t_2}{3} \][/tex]

Substitute [tex]\( t_2 = \frac{1}{2} \)[/tex],
[tex]\[ t_3 = \frac{\frac{1}{2}}{3} = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \][/tex]

Therefore, the value of [tex]\( t_3 \)[/tex] is [tex]\( \frac{1}{6} \)[/tex], which corresponds to option 2.