Given:
The sequence is
an=3n1+6n
. (1)
Calculation:
Obtain the first ten terms of the sequence.
Substitute 1 for n in equation (1).
a1=3⋅11+6⋅1 =37 =0.4286
Thus, the first term of the sequence is
a1=0.4286_
.
Substitute 2 for n in equation (1).
a2=3⋅21+6⋅2 =613 =0.4615
Thus, the second term of the sequence is
a2=0.4615_
.
Substitute 3 for n in equation (1).
a3=3⋅31+6⋅3 =919 =0.4737
Thus, the third term of the sequence is
a3=0.4737_
.
Substitute 4 for n in equation (1).
a4=3⋅41+6⋅4 =1225 =0.4800
Thus, the fourth term of the sequence is
a4=0.4800_
.
Substitute 5 for n in equation (1).
a5=3⋅51+6⋅5 =1531 =0.4839
Thus, the fifth term of the sequence is
a5=0.4839_
.
Substitute 6 for n in equation (1).
a6=3⋅61+6⋅6 =1837 =0.4865
Thus, the sixth term of the sequence is
a6=0.4865_
.
Substitute 7 for n in equation (1).
a7=3⋅71+6⋅7 =2143 =0.4884
Thus, the seventh term of the sequence is
a7=0.4884_
.
Substitute 8 for n in equation (1).
a8=3⋅81+6⋅8 =2449 =0.4898
Thus, the eighth term of the sequence is
a8=0.4898_
.
Substitute 9 for n in equation (1).
a9=3⋅91+6⋅9 =2755 =0.4909
Thus, the ninth term of the sequence is
a9=0.4909_
.
Substitute 10 for n in equation (1).
a10=3⋅101+6⋅10 =3061 =0.4918
Thus, the tenth term of the sequence is
a10=0.4839_
.
Therefore, the first ten terms of the sequence are tabulated below:
n 
an=3n1+6n

1 
0.4286 
2 
0.4615 
3 
0.4737 
4 
0.4800 
5 
0.4839 
6 
0.4865 
7 
0.4884 
8 
0.4898 
9 
0.4909 
10 
0.4918 
Plot the points
(n,an), for n=1,2,...10
on the graph as shown below in Figure 1.
Obtain the limit of the sequence (the value of the term
an
as n tends to infinity).
That is,
limn→∞an=limn→∞3n1+6n
.
Divide the numerator and the denominator by n.
limn→∞3n1+6n=limn→∞31n+6 =limn→∞3limn→∞1n+limn→∞6 =30+6 =12
Therefore, the limit of the sequence is
0.5_
.