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7/28/2019 Revision Proof by Induction
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Further Pure 1
Revision Topic 3: Proof by induction
The OCR syllabus says that candidates should be able to:(a) use the method of mathematical induction to establish a given result (not restricted to
summation of series);
(b) recognise situations where conjecture based on a limited trial followed by inductive proof is a
useful strategy, and carry this out in simple cases, e.g. to find the nth power of the matrix1 1
0 1
.
Section 1: General principals
There are two steps involved in proving a result by induction:
Step 1: Prove true when n = 1.
Step 2: (The inductive step). Assume the result is true forn= kand then prove true forn = k+ 1.
1.1 Summing series
Example: Prove by induction that2
61
( 1)(2 1)n
n
r
r n n=
= + + for all positive integer values ofn.
Solution: We wish to show that2 2 2 2
61 2 3 ... ( 1)(2 1)nn n n+ + + + = + + (*)
Step 1: This is to prove the result true when n = 1:
Left hand side of equation (*) = 12 = 1.
Right hand side of equation (*) =16
(1 1)(2 1)= + + = 1.
So equation (*) is true when n = 1.
Step 2: We assume the result is true when n = k, i.e. we assume that
2 2 2 2
61 2 3 ... ( 1)(2 1)kk k k+ + + + = + +
We want to prove the result is true when n = k + 1, i.e. we wish to show that
2 2 2 2 2 1
6
1 2 3 ... ( 1) ( 1 1)(2( 1) 1)kk k k k ++ + + + + + = + + + +
i.e.2 2 2 2 2 1
61 2 3 ... ( 1) ( 2)(2 3)kk k k k ++ + + + + + = + + .
But2 2 2 2 2 2
61 2 3 ... ( 1) ( 1)(2 1) ( 1)kk k k k k + + + + + + = + + + + (using our assumption).
So, [ ]( 1)2 2 2 2 2
61 2 3 ... ( 1) (2 1) 6( 1)
kk k k k k
++ + + + + + = + + +
Therefore,( 1) ( 1)2 2 2 2 2 2 2
6 61 2 3 ... ( 1) 2 6 6 (2 7 6)
k kk k k k k k k
+ + + + + + + + = + + + = + +
Factorising we get:2 2 2 2 2 1
6
1 2 3 ... ( 1) ( 2)(2 3)kk k k k ++ + + + + + = + + as required.
Therefore the result is true forn = k+ 1.
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So, by induction, the result is true for all integers n 1.
Examination style question
Prove by induction that 21
( 1)n
n
r
r n=
= + for all positive integer values ofn.
Worked examination question (AQA January 2006)
a) Prove by induction that
2 12 (3 2) (4 2 ) ... ( 1)2 2n nn n+ + + + + = (*)
for all integers n 1.
b) Show that
21 1
1
( 1)2 2 (2 1)n
r n n
r n
r n +
= ++ =
Solution:
Step 1: This is to prove the result true when n = 1:
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Left hand side of equation (*) = 1 1(1 1)2 2+ =
Right hand side of equation (*) = 1 21 = 2.
So equation (*) is true when n = 1.
Step 2: We assume the result is true when n = k, i.e. we assume that
2 12 (3 2) (4 2 ) ... ( 1)2 2k kk k+ + + + + = .
We want to prove the result true when n = k+ 1, i.e. we want to show that
2 1 1 12 (3 2) (4 2 ) ... ( 1 1)2 ( 1)2k kk k+ ++ + + + + + = +
i.e. that 2 1 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 ( 1)2k k kk k k ++ + + + + + + = + .
But, 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 ( 2)2k k k k k k k k + + + + + + + = + + (using the assumption)
So 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 ( ( 2))k k kk k k k + + + + + + + = + +
i.e. 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 (2 2)k k kk k k+ + + + + + + = +
i.e. 2 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 2( 1)k k kk k k+ + + + + + + = +
So we have 2 1 12 (3 2) (4 2 ) ... ( 1)2 ( 2)2 2 ( 1)k k kk k k ++ + + + + + + = + as required.
Therefore the equation is true when n = k+ 1.
So the equation is true for all integer values n 1.
b)2 2
1 1 1
1 1 1
( 1)2 ( 1)2 ( 1)2n n n
r r r
r n r r
r r r
= + = =
+ = + +
Using the result from part (a), we know that2
1 2
1
( 1)2 2 2n
r n
r
r n
=+ =
and1
1
( 1)2 2n
r n
r
r n
=+ = .
Therefore,2
1 2
1
( 1)2 2 2 2 2 2 2 2n
r n n n n n
r n
r n n n n
= ++ = =
Factorising gives:
21 1
1( 1)2 2 (2 2 1) 2 (2 1)
nr n n n n
r nr n n
+
= + + = = as required.
Examination question (OCR January 2005)
Prove by induction that13
1 4 2 5 3 6 ... ( 3) ( 1)( 5)n n n n n + + + + + = + +
for all integers n 1.
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Examination question (Edexcel)
Prove, by induction, that
2 112
1
( 1) ( 1)( 1)(3 2)n
r
r r n n n n=
= + +
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1.2 Sequences
Example: Prove that if 1 3 4 for n nu u n+ = + , and 1 2u = , then 14 3 2nnu
= .
Solution:
Step 1: We prove the formula true when n = 1: 1 11 4 3 2 4 2 2u= = = (which is true).
Step 2: Assume the formula is true when n = k, i.e. that 14 3 2kku
= .
We need to prove the result true when n = k+ 1, i.e. that 1 1
1 4 3 2 4 3 2
k k
ku
+
+ = = .
But,1
1 3 4 3 4 3 2 4k
k ku u
+ = + = + (using the assumption).
So, 11 3 4 3 6 4 4 3 2k k
ku
+ = + = as required.
Therefore the formula is true when n = k+ 1.
Therefore the result is true for all integers n 1.
Examination question (AQA June 2005)
The sequence 1 2 3, , ,...u u u is defined by
11 1 2
0, ( ).n nu u u n+= = +
Prove by induction that, for all n 1,
( )1
12
2.n
nu n
= +
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Worked examination question (Edexcel 2005)
(a) Express
3
106
+
+
x
xin the formp +
3+x
q, wherep and q are integers to be found.
(1)
The sequence of real numbers u1, u2, u3, ... is such that u1 = 5.2 and un + 1 =3
106
+
+
n
n
u
u
.
(b) Prove by induction that un > 5, forn+.(4)
Solution:
a) Note that p +3+x
q=
( 3)
3
p x q
x
+ +
+
.
Writing3
106
+
+
x
x=
( 3)
3
p x q
x
+ +
+
, we must have 6x + 10 =p(x + 3) + q.
So 6 =p (comparing coefficients ofx)
and 10 = 3p + q i.e. 10 = 18 + q i.e. q = -8.
Therefore3
106
+
+
x
x= 6 -
8
3x +.
b) un + 1 =3
106
+
+
n
n
u
u
= 6 -8
3nu +.
Step 1: Prove the result true when n = 1, i.e. that u1 > 5. This is trivially true as u1 = 5.2.
Step 2: Assume true when n = k, i.e. that uk> 5.
We then need to prove the result true when n = k+ 1, i.e. that uk+1 > 5.
As uk> 5, then uk+ 3 > 8 and so8
3nu +< 1.
Therefore
un + 1 =3
106
+
+
n
n
u
u
= 6 -8
3nu +> 6 1 = 5 (as required)
So the result is true when n = k+ 1.
Therefore the result is true forn+.
Examination question (NICCEA)
Consider the sequence defined by the relationship 1 5 2n nu u+ = + whose first term is 1 1u = .
(i) Show that the first four terms are 1, 7, 37, 187,
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(ii) Use the method of induction to prove that11
23(5 ) 1nnu
= .
1.3 Divisibility
Some questions give you a formula that defines a sequence (in the form un =f(n) ) and then ask
you to prove that all terms of the sequence are divisible by a particular number.
These questions are usually tackled by simplifyingf(k+ 1) f(k) ORf(k+ 1) f(k).
Adapted past examination question (adapted MEI):
Letf(n) = 4 12 3n+ + . By consideringf(n + 1) f(n), or otherwise, prove that 4 12 3n+ + is a
multiple of 5 for any positive integern.
Solution: Letf(n) = 4 12 3n+ + .
Step 1: Provef(1) is a multiple of 5. Butf(1) = 52 3+ = 35 (which is a multiple of 5).
Step 2: We assumef(k) is a multiple of 5. We want to prove thatf(k+ 1) is a multiple of 5.
First we try to simplify:
f(n + 1) -f(n) = 4( 1) 12 3n+ + + - ( 4 12 3n+ + ) = 4 5 4 12 2n n+ + = 4 1 4 4 12 (2 1) 15 2n n+ + =
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Therefore:
f(k+ 1) f(k) = 4 115 2 k+
So, f(k+ 1) = 4 115 2 ( )k f k+ + .
So,f(k+ 1) is a multiple of 5.
Therefore by induction,f(n) must be a multiple of 5 for all positive integers n.
Worked Examination Question: Edexcel 2002
Forn + prove that
23n + 2 + 5n + 1 is divisible by 3,
Solution:
Let f(n) = 23n + 2 + 5n + 1
If we here calculate, f(n + 1) +f(n) = 23(n+ 1) + 2 + 5n+ 1 + 1 + (23n + 2 + 5n + 1)
= 23n + 5 + 5n + 2 + 23n + 2 + 5n + 1
= 23n + 5 + 23n + 2 + 5n + 2 + 5n + 1
= 23n+2(23 + 1) + 5n+1(5 + 1)
= 9 23n+2 + 6 5n+1
Now we are ready to prove the result.
Step 1: First prove true when n = 1:
23 + 2 + 51 + 1 = 32 + 25 = 57 (which is divisible by 3).
Step 2: Assume that the result is true when n = k, i.e. thatf(k) is divisible by 3.
We showed above that f(k+ 1) +f(k) = 9 23k+2 + 6 5k+1
i.e. that f(k+ 1) = 9 23k+2
+ 6 5k+1
f(k)
So f(k+ 1) must be divisible by 3 (as required).
Therefore the result is true forn +
Past examination question (AQA January 2004)
The functionfis given by 3 3 3( ) ( 1) ( 2)f n n n n= + + + + .a) Simplify, as far as possible,f(n + 1) f(n).
Divisible by3 Divisible by3 Divisible by3 (byassumption)
Multiple of 5 Assumed to be a
multiple of 5
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b) Prove by induction that the sum of the cubes of three consecutive positive integers is divisible
by 9.
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Past examination question (Edexcel 2003)
f(n) = (2n + 1)7n 1.
Prove by induction that, for all positive integers n,f(n) is divisible by 4.
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Past examination question (Edexcel)
f(n) = 4 424 2 3n n + , where n is a non-negative integer.
a) Write down f(n + 1) f(n).
b) Prove by induction that f(n) is divisible by 5.
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1.4 Matrix results
Example:
a) Find the matrices
2 31 1 1 1
and0 1 0 1
.
b) Predict the value of1 1
0 1
n
and prove your result true by induction.
Solution
a)2
1 1 1 1 1 1 1 2
0 1 0 1 0 1 0 1
= = 3
1 1 1 2 1 1 1 3
0 1 0 1 0 1 0 1 = =
.
b) It seems sensible to predict that1 1 1
0 1 0 1
nn =
.
We know the result is true when n = 1, 2 and 3.
Suppose now that it is true when n = k, i.e. that
1 1 1
0 1 0 1
kk
= .
We need to prove the result true when n = k+ 1, i.e. that1
1 1 1 1
0 1 0 1
kk
+ + =
.
But:1
1 1 1 1 1 1 1 1 1
0 1 0 1 0 1 0 1 0 1
k kk
+ = =
(by assumption)
Therefore:1
1 1 1 1
0 1 0 1
kk
+ + =
(as required).
So the result is true forn = k+ 1.Therefore our prediction is true for all positive integer values ofn.
Past examination question (MEI adapted)
You are given the matrix1 4
1 3
=
A .
(i) Calculate 2 3andA A .
(ii) Prove by induction that1 2 4
1 2n n n
n n
= + A when n is a positive integer.
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Past examination question (Edexcel 2002)
Prove thatn
49
12=
+
139
31
nn
nn
when n is a positive integer.
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Past examination question
Let1 0
1 2A
= .
Use induction to prove that, for all positive integers n,
21 0
1 2 2
nnA =
.
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1.5 Other examination questions
Past examination question (OCR 2004)
(i) Show that 1 1 ( ) ( )k k k k k x y x x y y x y+ + = + .
(ii) Using this result, prove by induction that (x y) is a factor of n nx y for all integers n 1.
Past examination question (OCR)
Prove by mathematical induction that, for all positive integers n,
1 1 1 1 1... 1
2 4 8 2 2n n
+ + + + =