   Chapter 11.1, Problem 21E

Chapter
Section
Textbook Problem

# Calculate, to four decimal places, the first ten terms of the sequence and use them to plot the graph of the sequence by hand. Does the sequence appear to have a limit? If so, calculate it. If not, explain why.21. a n = 1 + ( − 1 2 ) n

To determine

To find: The first ten terms of the sequence and plot them on the graph to obtain the limit.

Explanation

Given:

The sequence is an=1+(12)n . (1)

Theorem used:

If limn|an|=0 , then limnan=0 . (2)

Calculation:

Obtain the first ten terms of the sequence.

Substitute 1 for n in equation (1).

a1=1+(12)1   =112   =12   =0.5

Thus, the first term of the sequence is a1=0.5000_ .

Substitute 2 for n in equation (1).

a2=1+(12)2   =1+14   =54   =1.25

Thus, the second term of the sequence is a2=1.2500_ .

Substitute 3 for n in equation (1).

a3=1+(12)3   =118   =78   =0.875

Thus, the third term of the sequence is a3=0.8750_ .

Substitute 4 for n in equation (1).

a4=1+(12)4   =1+116   =1716   =1.0625

Thus, the fourth term of the sequence is a4=1.0625_ .

Substitute 5 for n in equation (1).

a5=1+(12)5   =1132   =3132   =0.96875

Thus, the fifth term of the sequence is a5=0.9687_ .

Substitute 6 for n in equation (1).

a6=1+(12)6   =1+164   =6564   =1.015625

Thus, the sixth term of the sequence is a6=1.0156_ .

Substitute 7 for n in equation (1).

a7=1+(12)7   =11128   =127128   =0.9921875

Thus, the seventh term of the sequence is a7=0.9922_ .

Substitute 8 for n in equation (1).

a8=1+(12)8   =1+1256   =257256   =1.00390625

Thus, the eighth term of the sequence is a8=1.0039_ .

Substitute 9 for n in equation (1).

a9=1+(12)9   =11512   =511512   =0.998046875

Thus, the ninth term of the sequence is a9=0.9980_ .

Substitute 10 for n in equation (1).

a10=1+(12)10   =1+11024   =10251024   =1.0009765625

Thus, the tenth term of the sequence is a10=1.0010_ .

Therefore, the first ten terms of the sequence are tabulated below:

 n an=1+(−12)n 1 0

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