Page 35 - Schaum's Outline of Theory and Problems of Advanced Calculus

26 SEQUENCES [CHAP. 2

Solved Problems

SEQUENCES
2.1. Write the first five terms of each of the following sequences.

2n 1
3n þ 2
ðaÞ
n
1 ð 1Þ
n
ðbÞ 3
( )
n 1
ð 1Þ
ðcÞ
2 4 6 2n

1 1 1 1
ðdÞ þ þ þ þ n
2 4 8 2
( )
n 1 2n 1
ð 1Þ x
ðeÞ
ð2n 1Þ!
1 3 5 7 9
; ; ; ;
5 8 11 14 17
ðaÞ
2 2 2
ðbÞ 3 ; 0; 3 ; 0; 3
1 3 5
1 1 1 1 1
; ; ; ;
ðcÞ
2 2 4 2 4 6 2 4 6 8 2 4 6 8 10
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
ðdÞ ; þ ; þ þ ; þ þ þ ; þ þ þ þ
2 2 4 2 4 8 2 4 8 16 2 4 8 16 32
x x 3 x 5 x 7 x 9
; ; ; ;
1! 3! 5! 7! 9!
ðeÞ
Note that n! ¼ 1 2 3 4 n. Thus 1! ¼ 1, 3! ¼ 1 2 3 ¼ 6, 5! ¼ 1 2 3 4 5 ¼ 120, etc. We define
0! ¼ 1.

2.2. Two students were asked to write an nth term for the sequence 1; 16; 81; 256; ... and to write the
4
5th term of the sequence. One student gave the nth term as u n ¼ n . The other student, who did
2
3
not recognize this simple law of formation, wrote u n ¼ 10n 35n þ 50n 24. Which student
gave the correct 5th term?
4
4
4
4
4
If u n ¼ n ,then u 1 ¼ 1 ¼ 1, u 2 ¼ 2 ¼ 16, u 3 ¼ 3 ¼ 81, u 4 ¼ 4 ¼ 256, which agrees with the first four
4
terms of the sequence. Hence the first student gave the 5th term as u 5 ¼ 5 ¼ 625:
3
2
If u n ¼ 10n 35n þ 50n 24, then u 1 ¼ 1; u 2 ¼ 16; u 3 ¼ 81; u 4 ¼ 256, which also agrees with the first
four terms given. Hence, the second student gave the 5th term as u 5 ¼ 601:
Both students were correct. Merely giving a finite number of terms of a sequence does not define a
unique nth term. In fact, an infinite number of nth terms is possible.