Discussions on a mathematical induction problem
Yue Kwok Choy
Question
A sequence of real numbers {an} is defined as follows.
a 0  a 1  1,
a2  3
and
a n 3  3a n  2  a n 1  2a n
(a)
Let
,
n = 0, 1, 2, ….
b k  a k  2  a k 1  a k ,
k = 0, 1, 2, ….
b n  2b n 1
Prove that
(b)
for all n 
.
a n  2n .
Hence, or otherwise, deduce that
"Proposed" solution
(a)
Let P(n) be the proposition : " b n  2b n 1 " .
b0  a 2  a1  a 0  3  1  1  1
For P(1),
b1  a 3  a 2  a 1  (3a 2  a 1  2a 0 )  a 2  a 1  (3  3  1  2  1)  3  1  2

b1  2 b 0 .
Assume P(k) is true for some k 
,
b k  2b k 1
that is ,
….
(1)
….
(2)
For P(k + 1),
b k 1  a k 3  a k  2  a k 1
 3a k  2  a k 1  2a k  a k  2  a k 1
 2a k  2  a k 1  a k 
 2b k

P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true  n 
(b)
Let P(n) be the proposition : " a n  2
For P(0),
a0  1  2
For P(1),
a1  1  21
For P(2),
a 2  1  22 
n
.
".
0
P(0) , P(1), P(2) are true.
Assume P(k), P(k + 1) and P(k + 2) are true for some k 
k 1
ak  2 ,
a k1  2 ,
k
a k 2  2
, that is,
k 2
For P(k + 3) ,
a k 3  3a k  2  a k 1  2a k
(Note: we cannot use (2) directly as there are subtractions)
 3a k  2  a k 1  a k   2a k 1  a k
 3b k  2a k 1  a k
 32b k 1   2a k 1  a k
, by (a)
 34b k 2   2a k 1  a k
, by (a)
= ….
 
 32  1  22   2
 3 2k b0  2a k1  a k
k
k 1
k
 2 k 3  22  1
 2 k 3

P(k + 3) is true.
By the Second Principle of Mathematical Induction, P(n) is true
n
 {0}.
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Discussions
In the above "Proposed" solution,
(1)
The mathematical induction in (a) is wrong, as the inductive hypothesis ( b k  2b k 1 ) is not used in
proving P(k + 1) part. Therefore either you forfeit the use of mathematical induction and prove
directly as in Method 1 below or modify your mathematical induction as in Method 2 below.
Method 1
b n  a n  2  a n 1  a n
 3a n 1  a n  2a n 1   a n 1  a n
 2a n 1  a n  a n 1 
 2b n 1
Method 2
Let P(n) be the proposition : " b n  2b n 1 " .
b0  a 2  a1  a 0  3  1  1  1
For P(1),
b1  a 3  a 2  a 1  (3a 2  a 1  2a 0 )  a 2  a 1  (3  3  1  2  1)  3  1  2

b1  2 b 0 .
Assume P(k) is true for some k 
,
that is ,
b k  2b k 1
….
(1)
For P(k + 1),
By (1),
b k  2b k 1
a k  2  a k 1  a k  2a k 1  a k  a k 1 
2a k  2  2a k 1  2a k  22a k 1  2a k  3a k 1 
3a k 2  a k 1  2a k   a k 2  a k 1  23a k 1  a k  2a k 1   a k 1  a k 
2a k 3  a k  2  a k 1   22a k  2  a k 1  a k 
a k 3  a k  2  a k 1  2a k  2  a k 1  a k 
b k 1  2b k

P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true  n 
.
As can be seen, Method 1 is better than Method 2.
(2)
The Proposed solution in Part (b) is done better than part (a) but is not satisfactory:
(i)
The deduction in the P(k + 3) part as denoted by " = …. " is not wrong , but rather
uncomfortable. You may better write
" = ….
(use deduction, hence induction)" to tell others
that you know how to use induction here, but you are not writing for simplicity.
(ii)
P(k + 2) , that is , a k  2  2k  2
is not really used in the proof of P(k + 3) in the above.
You may cut that part, but it is still all right to leave it there.
You can in fact prove separately b n  2 n by mathematical induction (or deduction) first and the
proof of part (b) can be rewritten as follows:
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Let P(n) be the proposition : " a n  2 n " .
For P(0),
a 0  1  20
For P(1),
a1  1  21
For P(2),
a 2  1  22 
P(0) , P(1), P(2) are true.
Assume P(k) and P(k + 1) are true for some k 
ak  2 ,
k
a k1  2
k 1
, that is,
….
(2)
For P(k + 3) ,
a k 3  3a k  2  a k 1  2a k
 3a k  2  a k 1  a k   2a k 1  a k
 3b k  2a k 1  a k
 3  2 k  2  2 k 1  2 k
 3  2k  4  2k  2k
 8  2k
 2 k 3

P(k + 3) is true.
By the Second Principle of Mathematical Induction, P(n) is true  n 
 {0}.
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