From the given general terms, find the first 5 terms and present them in the form of a sequence when [tex]\( n \)[/tex] represents a natural number:

(a) [tex]\( t_n = 2n + 4 \)[/tex]

(b) [tex]\( t_n = 3n - 1 \)[/tex]

(c) [tex]\( t_n = 3^n \)[/tex]

(d) [tex]\( t_n = n^2 - 1 \)[/tex]

(e) [tex]\( t_n = (-1)^n \cdot n^2 \)[/tex]

(f) [tex]\( t_n = n^2 + 2n + 3 \)[/tex]

(g) [tex]\( t_n = 3n^2 - 5 \)[/tex]



Answer :

Sure! Let's find the first 5 terms of each sequence step-by-step, where [tex]\( n \)[/tex] represents the natural numbers [tex]\( 1, 2, 3, 4, 5 \)[/tex].

(a) For [tex]\( t_n = 2n + 4 \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= 2(1) + 4 = 6, \\ t_2 &= 2(2) + 4 = 8, \\ t_3 &= 2(3) + 4 = 10, \\ t_4 &= 2(4) + 4 = 12, \\ t_5 &= 2(5) + 4 = 14. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( 6, 8, 10, 12, 14 \)[/tex].

(b) For [tex]\( t_n = 3n - 1 \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= 3(1) - 1 = 2, \\ t_2 &= 3(2) - 1 = 5, \\ t_3 &= 3(3) - 1 = 8, \\ t_4 &= 3(4) - 1 = 11, \\ t_5 &= 3(5) - 1 = 14. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( 2, 5, 8, 11, 14 \)[/tex].

(c) For [tex]\( t_n = 3^n \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= 3^1 = 3, \\ t_2 &= 3^2 = 9, \\ t_3 &= 3^3 = 27, \\ t_4 &= 3^4 = 81, \\ t_5 &= 3^5 = 243. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( 3, 9, 27, 81, 243 \)[/tex].

(d) For [tex]\( t_n = n^2 - 1 \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= 1^2 - 1 = 0, \\ t_2 &= 2^2 - 1 = 3, \\ t_3 &= 3^2 - 1 = 8, \\ t_4 &= 4^2 - 1 = 15, \\ t_5 &= 5^2 - 1 = 24. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( 0, 3, 8, 15, 24 \)[/tex].

(e) For [tex]\( t_n = (-1)^n \cdot n^2 \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= (-1)^1 \cdot 1^2 = -1, \\ t_2 &= (-1)^2 \cdot 2^2 = 4, \\ t_3 &= (-1)^3 \cdot 3^2 = -9, \\ t_4 &= (-1)^4 \cdot 4^2 = 16, \\ t_5 &= (-1)^5 \cdot 5^2 = -25. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( -1, 4, -9, 16, -25 \)[/tex].

(f) For [tex]\( t_n = n^2 + 2n + 3 \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= 1^2 + 2(1) + 3 = 6, \\ t_2 &= 2^2 + 2(2) + 3 = 11, \\ t_3 &= 3^2 + 2(3) + 3 = 18, \\ t_4 &= 4^2 + 2(4) + 3 = 27, \\ t_5 &= 5^2 + 2(5) + 3 = 38. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( 6, 11, 18, 27, 38 \)[/tex].

(g) For [tex]\( t_n = 3n^2 - 5 \)[/tex]:
[tex]\[ \begin{aligned} t_1 &= 3(1)^2 - 5 = -2, \\ t_2 &= 3(2)^2 - 5 = 7, \\ t_3 &= 3(3)^2 - 5 = 22, \\ t_4 &= 3(4)^2 - 5 = 43, \\ t_5 &= 3(5)^2 - 5 = 70. \end{aligned} \][/tex]

So, the first 5 terms are [tex]\( -2, 7, 22, 43, 70 \)[/tex].

Overall, the sequences are:

(a) [tex]\( 6, 8, 10, 12, 14 \)[/tex]

(b) [tex]\( 2, 5, 8, 11, 14 \)[/tex]

(c) [tex]\( 3, 9, 27, 81, 243 \)[/tex]

(d) [tex]\( 0, 3, 8, 15, 24 \)[/tex]

(e) [tex]\( -1, 4, -9, 16, -25 \)[/tex]

(f) [tex]\( 6, 11, 18, 27, 38 \)[/tex]

(g) [tex]\( -2, 7, 22, 43, 70 \)[/tex]

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