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

2 Answers
Jan 14, 2018

#-5,-4, 9, 40, 95#

Explanation:

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

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

#a_n=n(n^2-6)#

So,

#a_1=1(1-6)=-5# (first term)

#a_2=2*(4-6)=-4# (second term)

#a_3=3(9-6)=9# (third term)

#a_4=4(16-6)=40# (fourth term)

#a_5=5(25-6)=95# (fifth term)

Hence, the first five terms of the sequence #a_n=n(n^2-6)# are

#-5,-4, 9, 40, 95#

Jan 14, 2018

#-5,-4,9,40,95#

Explanation:

#"substitute corresponding value for n into "a_n=n(n^2-6)#

#a_n=n(n^2-6)=n^3-6n#

#rArra_(color(red)(1))=(color(red)(1))^3-(6xxcolor(red)(1))=1-6=-5#

#rArra_(color(red)(2))=(color(red)(2))^3-(6xxcolor(red)(2))=8-12=-4#

#rArra_(color(red)(3))=(color(red)(3))^3-(6xxcolor(red)(3))=27-18=9#

#rArra_(color(red)(4))=(color(red)(4))^3-(6xxcolor(red)(4))=64-24=40#

#rArra_(color(red)(5))=(color(red)(5))^3-(6xxcolor(red)(5))=125-30=95#