How do you write the first five terms of the sequence $a_{n} = n (n^{2} - 6)$ $a_n=n(n^2-6)$ ?

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How do you write the first five terms of the sequence $a_{n} = n (n^{2} - 6)$ $a_n=n(n^2-6)$ ?

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Answer 1 · 2018-01-14T09:11:08

$- 5, - 4, 9, 40, 95$ $-5,-4, 9, 40, 95$

Explanation:

Just plug in $n = 1, 2, 3, 4, 5$ $n=1,2,3,4,5$ for the formula to get the first five terms.

We know that the $n$ $n$ th term of the sequence is defined by

$a_{n} = n (n^{2} - 6)$ $a_n=n(n^2-6)$

So,

$a_{1} = 1 (1 - 6) = - 5$ $a_1=1(1-6)=-5$ (first term)

$a_{2} = 2 \cdot (4 - 6) = - 4$ $a_2=2*(4-6)=-4$ (second term)

$a_{3} = 3 (9 - 6) = 9$ $a_3=3(9-6)=9$ (third term)

$a_{4} = 4 (16 - 6) = 40$ $a_4=4(16-6)=40$ (fourth term)

$a_{5} = 5 (25 - 6) = 95$ $a_5=5(25-6)=95$ (fifth term)

Hence, the first five terms of the sequence $a_{n} = n (n^{2} - 6)$ $a_n=n(n^2-6)$ are

$- 5, - 4, 9, 40, 95$ $-5,-4, 9, 40, 95$

Answer 2 · 2018-01-14T09:18:37

$- 5, - 4, 9, 40, 95$ $-5,-4,9,40,95$

Explanation:

$substitute corresponding value for n into a_{n} = n (n^{2} - 6)$ $"substitute corresponding value for n into "a_n=n(n^2-6)$

$a_{n} = n (n^{2} - 6) = n^{3} - 6 n$ $a_n=n(n^2-6)=n^3-6n$

$\Rightarrow a_{1} = {(1)}^{3} - (6 \times 1) = 1 - 6 = - 5$ $rArra_(color(red)(1))=(color(red)(1))^3-(6xxcolor(red)(1))=1-6=-5$

$\Rightarrow a_{2} = {(2)}^{3} - (6 \times 2) = 8 - 12 = - 4$ $rArra_(color(red)(2))=(color(red)(2))^3-(6xxcolor(red)(2))=8-12=-4$

$\Rightarrow a_{3} = {(3)}^{3} - (6 \times 3) = 27 - 18 = 9$ $rArra_(color(red)(3))=(color(red)(3))^3-(6xxcolor(red)(3))=27-18=9$

$\Rightarrow a_{4} = {(4)}^{3} - (6 \times 4) = 64 - 24 = 40$ $rArra_(color(red)(4))=(color(red)(4))^3-(6xxcolor(red)(4))=64-24=40$

$\Rightarrow a_{5} = {(5)}^{3} - (6 \times 5) = 125 - 30 = 95$ $rArra_(color(red)(5))=(color(red)(5))^3-(6xxcolor(red)(5))=125-30=95$

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How do you write the first five terms of the sequence $a_{n} = n (n^{2} - 6)$ $a_n=n(n^2-6)$ ?

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How do you write the first five terms of the sequence $a_{n} = n (n^{2} - 6)$ $a_n=n(n^2-6)$ ?