Download Mathematical Induction (selected questions) 1. (a) Let P(n) be the

Mathematical Induction (selected questions)
1.
(a)
Let P(n) be the proposition :
For P(1),
1 + 2 + ⋯ + n = nn + 1
L.H.S.=1 = × 1 × 1 + 1
=R.H.S.,
∴P(1) is true.
Assume P(k) is true for some natural number k, that is, 1 + 2 + ⋯ + k = kk + 1
………(1)
For P(k + 1),
1 + 2 + ⋯ + k + k + 1
= kk + 1
+ k + 1
, by (1)
= k + 1
k + 2
= k + 1
k + 1
+ 1
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true for all natural numbers n.
1.
(b)
Let P(n) be the proposition :
For P(1),
1 + 2 + ⋯ + n = nn + 1
2n + 1
L.H.S.=1 = 1 = × 1 × 2 + 1
2 × 1 + 1
=R.H.S.,
∴ P(1) is true.
Assume P(k) is true for some k∈ N, that is, 1 + 2 + ⋯ + k = kk + 1
2k + 1
………(1)
For P(k + 1),
1 + 2 + ⋯ + k + k + 1
= kk + 1
2k + 1
+ k + 1
,
by (1)
= k + 1
k2k + 1
+ 6k + 1
= k + 1
2k + 7k + 6
= k + 1
k + 2
2k + 3
= k + 1
k + 1
+ 12k + 1
+ 1
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(c)
Let P(n) be the proposition : “x2n – y2n is divisible by x + y for any integers x, y and positive integer n.”
or “x − y = x + y
g x, y
, where g x, y
is a polynomial in x ,y and x, y ∈ Z, n∈N.
For P(1),
x − y = x + y
x − y
= x + y
g x, y
,
∴ P(1) is true.
Assume P(k) is true for some k∈ N, that is,
x − y = x + y
g x, y
, where g x, y
∈ Z[x, y]
For P(k + 1),
x − y = x − y x + y = x + y
g x, y
x + y = x + y
g x, y
, where g x, y
= g x, y
x + y ∈x, y
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(d)
Let P(n) be the proposition :
For P(1),
1 + 2 + ⋯ + n =
L.H.S.=1 = 1 =
nn + 1
!
× 1 × 1 + 1
! =R.H.S. ,
Assume P(k) is true for some k∈ N, that is, 1 + 2 + ⋯ + k =
∴ P(1) is true.
kk + 1
! ………(1)
1
For P(k + 1),
1 + 2 + ⋯ + k + k + 1
=
kk + 1
! + k + 1
,
by (1)
= k + 1
k + 4k + 1
"
= k + 1
k + 4k + 4
"
= k + 1
k + 1
"
= $ k + 1
k + 1
+ 1%
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(e)
Let P(n) be the proposition :
For P(1),
2 × 1 + 3 × 2 + 4 × 2 + 5 × 2 + ⋯ + n + 1
2' = n2
2 × 1 = 2 = 1 × 2 ,
∴
P(1) is true.
Assume P(k) is true for some k∈ N , that is, 2 × 1 + 3 × 2 + 4 × 2 + 5 × 2 + ⋯ + k + 1
2' = k2 … 1
For P(k + 1),
2 × 1 + 3 × 2 + 4 × 2 + 5 × 2 + ⋯ + k + 1
2' + k + 2
2
= k2 + k + 2
2
, by (1)
= k + k + 2
2 = 2k + 1
2 = k + 1
2 .
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(f)
Let P(n) be the proposition :
For P(1),
1− = =
1− + − +⋯−
"
∴
,
=
+
+
P(1) is true.
Assume P(k) is true for some k∈ N , that is, 1 − + − + ⋯ −
For P(k + 1),
=
∴
.
1 − + − + ⋯ −
+
"
+ ⋯+
+
+
−
+
"
−
=
+
+
+ ⋯+
=
+
+ ⋯+
+
+ ⋯+
… 1
! + − , by (1)
! = + + ⋯ + + + P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(g)
Let P(n) be the proposition : 2" − 1 = 15a
For P(1),
2" − 1 = 15 = 15a , where a = 1ϵ+.
Assume P(k) is true for some k∈ N , that is, 2
For P(k + 1),
2
"
− 1 = 16 × 2
"
"
∴
P(1) is true.
− 1 = 15a … 1
, where a ϵ+.
− 1 = 16 × -2" − 1. + 15 = 16 × 15a + 15 , by (1)
= 1516a + 1
= 15a,
∴
where a ϵ+.
where a ϵ+.
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(h)
Let P(n) be the proposition :
1
The no. of diagonals in a convex n-sided polygon is f(n)= n ( n − 3) (n ≥ 3).
For P(3), There is no diagonal in a triangle. f(3) = 0.
2
∴
P(3) is true.
2
Assume P(k) is true for some k∈ N \{1,2}, that is,
1
The no. of diagonals in a convex k-sided polygon is f(k)= k ( k − 3)
2
For P(k + 1),
By (1),
Let
(k – 1)
f(k) + (k – 1) =
1
P1P2….Pk is f(k)= k ( k − 3)
2
k-sided polygon
diagonals connected to the the point
The no. of diagonals in a convex
………(1)
be the vertices in clockwise direction of the (k+1)-sided polygon.
The no. of diagonals in a convex
There are additional
∴
P1, P2, …., Pk, Pk+1
(k ≥ 3)
(k+1)-sided polygon
Pk+1, namely, Pk+1P2, Pk+1P3, …, Pk+1Pk.
P1P2….PkPk+1
is
1
1
1
1
k ( k − 3) + ( k − 1) = {k ( k − 3) + 2( k − 1)} = {k 2 − k − 2} = {( k + 1)[( k + 1) − 3]}
2
2
2
2
∴
= f(k + 1).
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(i)
Let Pn
be the proposition:
m, n ≥ 2,
a1, a2 , …, an,
a + a + ⋯ + a
b + b + ⋯ + b1 =
For P(2),
b1, b2, …, bm ∈ R,
∑1
45 ∑35 a 3 b4
= ∑35 ∑1
45 a 3 b4
a + a
b + b + ⋯ + b1 = a b + a b + a b + a b + ⋯ + a b1 + a b1 = ∑35 a3 b + ∑35 a3 b + ⋯ + ∑35 a3 b1 = ∑1
45 ∑35 a 3 b4
Also,
a + a
b + b + ⋯ + b1 = a b + a b + ⋯ + a b1
+ ab + a b + ⋯ + a b1 1
1
= ∑1
45 a b4 + ∑45 a b4 = ∑35 ∑45 a 3 b4
∴
P(2) is true.
Assume P(k) is true for some k ∈ N\{1},
1
a + a + ⋯ + a
b + b + ⋯ + b1 = ∑1
45 ∑35 a 3 b4 = ∑35 ∑45 a 3 b4 … 1
For P(k + 1),
a + a + ⋯ + a + a b + b + ⋯ + b1 = a ′ + a ′ + ⋯ + a ′
b + b + ⋯ + b1 where a7 = a , … , a7' = a' , a7 = a + a
1
= ∑1
45 ∑35 a 3 ′b4 = ∑45a b4 + a b4 + ⋯ + a ' b4 + a + a b4 , by (1)
1
= ∑1
45a b4 + a b4 + ⋯ + a b4 + a b4 = ∑45 ∑35 a 3 b4
a + a + ⋯ + a + a b + b + ⋯ + b1 Also,
= a ′ + a ′ + ⋯ + a ′
b + b + ⋯ + b1 where a7 = a , … , a7' = a' , a7 = a + a
1
1
1
1
= ∑35 ∑1
45 a 3 ′b4 = ∑45 a b4 + ∑45 a b4 + ⋯ + ∑45 a ' b4 + ∑45a + a b4
1
1
1
= ∑1
45 a b4 + ∑45 a b4 + ⋯ + ∑45 a ' b4 + ∑45-a b4 + a b4 .
1
1
1
1
= ∑1
45 a b4 + ∑45 a b4 + ⋯ + ∑45 a ' b4 + + ∑45 a b4 + ∑45 a b4
1
= ∑
35 ∑45 a 3 b4
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N\{1}.
1.
(j)
3n-2 ≥ n5 for n ≥ 20.
Let P(n) be the proposition :
For P(20),
320-2
= 318 = (33)6 = 276 ≥
Assume P(k) is true for some k ∈ N ,
For P(k + 1),
3
(k+1)-2
= 3(3
k–2
where
5
) ≥ 3(k )
206 ≥ 205.
∴
k ≥ 20, that is, 3k-2 ≥ k5
P(20) is true.
………(1)
, by (1)
3
= k5 + 3k (k4)
≥
≥
≥
k5 + 0.4 k k4 + 0.4 k2k3 + 0.4 k3 k2 + 0.4 k4 k + 0.4 k5
k5 + 0.4(20)k4 + 0.4(20)2k3 + 0.4(20)3k2 + 0.4(20)4k + 0.4 (20)5,
5
4
3
k + 5k + 10k + 10k + 5k + 1
= (k + 1)5
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N,
1.
(k)
since k ≥ 20.
2
n ≥ 20.
2n > n2 for n ≥ 5.
Let P(n) be the proposition :
28 = 32 > 25 = 5
For P(5),
where
∴
P(5) is true.
Assume P(k) is true for some k ∈ N, where k ≥ 5, that is 2 > k … 1
For P(k + 1),
2 = 2-2 . > 2k = k + k ≥ k + 5k = k + 2k + 3k
≥ k + 2k + 35
> k + 2k + 1 = k + 1
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N,
1.
(l)
∑1
5
Let P(m) be the proposition :
For P(1),
…;
;!
=1=
…;'
;!
…;
;
Assume P(k) is true for some k ∈ N, that is ∑5
For P(k + 1),
∑
5
∴
;
!
…;'
;!
=
…;
;
!
=
…;
;
!
+
(m) Let P(n) be the proposition : C? =
For P(0),
…;
… 1
;
!
…;'
+
…;'
= ∑5
;!
;!
, by (1)
;!
…;
;
;
!
P(k + 1) is true.
∀ m ∈ N.
is always an integer, where 0 ≤ r ≤ n
CCC = 1 , which is an integer.
Assume P(k) is true for some k∈ +⋃{0}, that is, C? =
For P(k + 1),
CC = C
= 1 are integers.
For r ≥ 1,
C? = C? + C?'
∴
P(1) is true.
=
k + p + 1
=
!
n ≥ 5.
11
1
…1;
;
!
;!
…;
?!'?
!
P(k + 1) is true.
…;'
By the Principle of Mathematical Induction, P(m) is true
1.
=
where
∴
!
?!'?
!
is always an integer. …(1)
is the sum of two integers by (1), and is an integer.
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N ∪ {0}.
1.
(n)
Let P(n) be the proposition :
2×4 + 3 = 11a
, where a ∈ +
4
For
2×4 + 3" = 209 = 11 × 19 = 11a
P(1),
2×4
Assume P(k) is true for some k∈ N , that is,
For
P(k+1),
∴ P(1) is true.
+ 3 = 11a … 1
, where a ∈ +
2×4 + 3" = 32×4 + 27× 3
= 16-2×4 + 3 . + 11× 3
= 1611a + 11× 3
, by (1)
= 11-16a + 3 . = 11a
∴
, where a = 16a + 3 ∈ +.
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(o)
4 + 3 = 13a
Let P(n) be the proposition :
For
4 + 3 = 91 = 13×7 = 13a
P(1),
P(k+1),
∴
P(1) is true.
4 + 3 = 13a … 1
, where a ∈ +
Assume P(k) is true for some k∈ N , that is,
For
, where a ∈ + .
4 + 3 = 16×4 + 3× 3
= 3-4 + 3 . + 13 × 4
= 313a + 13 × 4
= 13-3a + 3
∴
, by (1)
, where a = 3a + 3 ∈ +
. = 13a
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
1.
(p)
Let P(n) be the proposition :
For
P(1),
+
= < =1−
+⋯+
P(k+1),
+
<1−
∴
+ ⋯+
+ J
<1−
−
=J
+
+
P(1) is true.
(We change the proposition and prove this first.)
∴
Assume P(k) is true for some k∈ N , that is,
For
+⋯+
+⋯+
<1−
… 1
K + J + − K , by (1)
K = 1 − = 1 − < 1 − P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
∴
1.
(q)
+
+ ⋯+
<1−
Let P(n) be the proposition :
is
1
1
2
4
f(n) = ( n + 1) +
<1
The number of pairs of non-negative integers (x, y) satisfying
x + 2y = n
[1 + ( −1) ]
n
For P(1),
(x, y) = (1, 0) ,
For P(2),
(x, y) = (2, 0)
f(1) = 1
and (0, 1) are solutions.
Also, f(2) = 2
∴
P(1) is true.
∴
P(2) is true.
Assume P(k) is true for some k∈ N, that is,
The number of pairs of non-negative integers (x, y) satisfying
x + 2y = k
is
5
1
1
f(k) = ( k + 1) + [1 + ( −1) k ]
2
4
For P(k + 2),
………(1)
Consider the equation
x + 2z = k
………(2)
1
1
The number of pairs of non-negative integers (x, z) satisfying (2) is f(k) = ( k + 1) + [1 + ( −1) k ] , by (1)
2
4
Put
z = y – 1,
(2) becomes
x + 2(y – 1) = k
Obviously (x, y) = (k + 2, 0) satisfies (3), but
or
x + 2y = k + 2
………(3)
(x, z) = (x, y – 1) = (k + 2, – 1) does not satisfy (2).
∴ The total number of pairs of non-negative integers (x, y) satisfying (3) is
∴
1
1+
1 + f(k) =
2
( k + 1) +
1
4
[1 + (−1) ] = 1 [(k + 2) + 1] + 1 [1 + (−1) ] = f (k + 2)
k +2
k
2
4
P(k + 2) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
2.
(a)
(
Let P(n) be the proposition :
For
P(0),
(
3 +1 −
For
P(1),
(
3 +1 −
Assume
(
)
3 +1
2 k −1
−
(
(
)
1
2 k −1
[(
)
3
and P(k)
2 n +1
= 2 n +1 a n , where an ∈ Z.
2 k +3
−
(
= 2 k +1 a k × 8 −
(
3 +1
−
(
)
3 −1
2
a0 = 1
]
)
2
(
)
3 +1
(k ≥ 1), that is,
k∈ N
2 k +1
−
(
)
3 −1
2 k +1
= 2 k +1 a k ….(*)
)
3 −1
2 k +1
)(
where
are true for some
= 2 k a k −1 ,
)
3 +1
2 k +1
)
3 −1
3 − 1 = 2 3 3 + 1 = 20 = 21+1 a1 , where a1 = 5.
[( 3 + 1)
=
(
) (
)
For P(k + 1),
−
3 − 1 = 2 = 2a 0 ,
3
3 −1
2 n +1
) (
1
P(k – 1)
)
3 +1
2 k +3
] [( 3 + 1) + ( 3 − 1) ]− ( 3 − 1) ( 3 + 1)
2
) [(
3 −1
2
2
)
3 +1
2 k −1
−
(
2
)
3 −1
2 k −1
2 k +1
+
(
)(
3 +1
2
)
3 −1
2 k +1
, by (*)
]
= 2k +2 a k × 4 − 8 × 2 k a k −1 = 2 k +2 ( 4a k − a k −1 ) = 2 k +2 a k +1 , where ak+1 = 4ak – ak-1∈ Z.
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N ∪ {0}.
2.
(b)
Let
α = 3 + √5, β = 3 − √5 , then α + β = 6, αβ = 4.
α + β = 2a , where
Let P(n) be the proposition :
For P(1), α + β = 6 = 2 a
an ∈ Z.
, where a1 =3 ∈ N.
For P(2), α + β = α + β
− 2αβ = 6 − 2 × 4 = 28 = 2 a
Assume P(k) and P(k + 1)
, where
a2 =7 ∈Z .
are true, that is,
α + β = 2 a ,α + β = 2 a … ∗
, where ak , ak+1 ∈ Z.
For P(k + 2), α + β = -α + β .α + β
− αβ
-α + β . = -2 a .6
− 4
-2 a .
6
= 2 3a − a = 2 a ,
∴
where a ∈.
P(k+2) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
3.
Let P(n) be the proposition :
There are two collections of marbles of the same quantity, n. According to the rules of taking
marbles set by the question, the first player wins.
For P(1), The first player take 1 marble from one collection and the second player takes the last marble and wins.
∴
P(1) is true.
Assume P(1), P(2), …, P(k) are true. So if there are two collections of marbles of the same quantity less than or equal to k, the
second player wins.
For P(k + 1),
There are two collections of marbles of the same quantity, k + 1.
If the first player takes away any number of marbles, say p, 1 ≤ p ≤ k + 1
from one collection. The second player
takes away also p marbles from the other collection. So now we have two collections of marbles each of k + 1– a ≤ k.
∴
Thus by the inductive hypothesis, the second player wins.
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
4.
(a)
(
Let P(n) be the proposition :
(
For P(2),
m2 + 1 − m
P(1) , P(2)
)
2
n
= a n m2 + 1 − bn ,
For P(k + 1),
a1 = 1,
where an , bn ∈ Z, m ∈ N.
b1 = m.
= 2 m 2 + 1 − 2 m m 2 + 1 = a 2 m 2 + 1 − b 2 , where a2 = – 2m ,
b2 = –(2m2 + 1).
are true.
Assume P(k) is true for some
(
m2 + 1 − m
k∈ N, that is,
)
k +1
=
(
)(
where
a k +1 = −(a k m + b k ),
= a k m2 + 1 − bk
∴
)
m 2 + 1 − m = a 1 m 2 + 1 − b1 , where
For P(1),
∴
m2 + 1 − m
(
(
m2 + 1 − m
m2 + 1 − m
)(
k
)
k
= a k m2 + 1 − bk
m2 + 1 − m
………(1)
)
)
m 2 + 1 − m = ( − a k m − b k ) m 2 + 1 − [− (a k ( m 2 + 1) + b k m )] = a k +1 m 2 + 1 − b k +1
b k +1 = − (a k m 2 + a k + b k m )
……….(2)
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
(b)
b n = − (a n −1m 2 + a n −1 + b n −1m )
From (2),
a n = −(a n −1m + b n −1 ),
From (3),
a n ( m 2 + 1) − b n = ( − a n −1m − b n −1 ) ( m 2 + 1) − (a n −1m 2 + a n −1 + b n −1m )
2
2
……….(3)
2
2
[
] = (−1) [1(m
]
[
]
= b n −1 − a n −1 ( m 2 + 1) = − a n −1 ( m 2 + 1) − b n −1 = ( − 1) a n −2 ( m 2 + 1) − b n −2 = ...
2
= ( −1)
n −1
[a (m
1
2
+ 1) − b1
2
n −1
2
2
2
+ 1) − m ] = ( −1)
2
2
n −1
……….(4)
From (3),
7
(i)
If
n
∴
(
(ii)
m2 + 1 − m
n
)
n
a n ( m 2 + 1) − b n = ( −1) n = −1
2
bn < 0.
2
∴
6.
an < 0,
a n ( m 2 + 1) + 1 = b n
If
(
is even,
2
2
……….(5)
= a n m 2 + 1 − b n = − b n − a n m 2 + 1 = b n − a n ( m 2 + 1) = N + 1 − N , by (5).
2
is odd,
an > 0,
2
a n ( m 2 + 1) − b n = ( −1) n = 1
2
bn > 0.
2
a n ( m 2 + 1) = b n + 1
2
2
m2 + 1 − m
)
n
……….(6)
= a n m 2 + 1 − b n = a n ( m 2 + 1) − b n = N + 1 − N , N∈ N
2
2
First, show by Math. Induction the proposition P(n) : 1 + 3 + … + (2n – 1) = n2 ∀n∈N
Then,
=1 =1
a2 = 3 + 5
= 8 = 23
a3 = 7 + 9 + 11
= 27 = 33
:
:
:
2
2
2
an = n2 + n – 1 = 2

n
Adding up all equalities,
∑
r =1
2
b1 = 1 (no. of term = 1 ),
The last term of
n ( n + 1) 
 − 1.
2 
2
n ( n + 1)    n ( n + 1) 
1 2
2
r 3 = 1 + 3 + 5 + ... + 2
 − 1 = 
 = n ( n + 1) , by (1).
 
2   
2 
4
b2 = 3 + 5 + 7 + 9 (no. of terms = 22),
b3 = 11 + 13 + 15 + … + 27 (no. of terms = 32),
1
bn = 2[12 + 22 + …+ n2] – 1 = 2 n ( n + 1)( 2 n + 1)  − 1
6

By (1),
1
Sn = b1 + b2 + … + bn =  n ( n + 1)( 2 n + 1) 
6

By (2) ,
The last term of bn-1 =
1
6
….(2)
2
….(3)
( n − 1)( n )( 2 n − 1) − 1 .
2
2
1
1
1
Sn – Sn-1 =  n ( n + 1)( 2 n + 1)  −  ( n − 1)( n )( 2 n − 1)  = [2 n 5 + n 3 ]
6
 6

3
Hence
bn =
Hence
Sn = b1 + b2 + … + bn =
1 n 5
r +
2
3  r =1
∑
n
Compare (3) and (4), we get
∑
r5 =
r =1
7.
:
= n3
an = (n – n + 1) + (n – n + 3) + … + [n – n + (2n – 1)]
The last term of
….(1)
3
a1 = 1
:
, by (6)
n
∑
r =1
 1 n 5 1 2

r 3  = 2
r + n ( n + 1) 2 
4
 3  r =1

∑
….(4)
2
1 1
1 2
1 2

2
2
2
3
n
(
n
+
1
)(
2
n
+
1
)
 − n ( n + 1)  = n ( n + 1) ( 2 n + 2 n − 1)
 

2 6
4
12

(Backward Mathematical Induction)
(i)
(a)
I(n) : If
For
xi ∈ [a, b],
 x + ... + x n 
i = 1, 2, …, n, then f ( x 1 ) + ... + f ( x n ) ≤ nf  1



n
I(21), since it is given that
 x + x2
f ( x 1 ) + f ( x 2 ) ≤ 2f  1
 2

.

∴ I(21)
is true.
8
 x 1 + ... + x 2 k 
is true. i.e. f ( x 1 ) + ... + f ( x 2 k ) ≤ 2 k f 

2k


f ( x 1 ) + ... + f ( x 2 k ) + f ( x 2 k +1 ) + ... + f ( x 2 k +1 )
Assume I(2k)
I(2k+1),
For
….(1)
 x 1 + ... + x 2 k 
 x k + ... + x 2 k +1 
  x 1 + ... + x 2 k   x 2 k +1 + ... + x 2 k +1
≤ 2k f 
 + 2 k f  2 +1 k
 = 2 k f 
 + f
k
2
2
2k
2k




 
 

 , by (1)

  x 1 + ... + x 2k   x 2 k +1 + ... + x 2 k +1  
  1  x 1 + ... + x 2 k x 2 k +1 + ... + x 2 k +1  
= 2 k f 
+
 + f
  ≤ 2 k 2 f  
  , by I(2)
k
k
2
2
2k
2k
 
 

 2
 
  x 1 + ... + x 2 k + x 2 k +1 + ... + x 2 k +1
= 2 k +1 f 
2 k +1
 
(b)



∴ I(2k+1)
is true.
is true (n ≥ 2),
Assume I(n)
i.e.
 x + ... + x n
f ( x 1 ) + ... + f ( x n ) ≤ nf  1

n
Put
xn =
x 1 + ... + x n −1
n −1
x 

 n − 1  x 1 + ... + x n −1
+ n 
 = nf 


 n 
n −1
n − 1
, then
 x + ... + x n −1 
 x 1 + ... + x n −1 
f ( x 1 ) + ... + f ( x n −1 ) + f  1
 ≤ nf 





n −1
n −1
x + ... + x n −1 
f ( x 1 ) + ... + f ( x n −1 ) ≤ ( n − 1)f  1



n −1
(c)
(ii)
∀n ∈ N,
∃ (k ∈ N and r ∈ N)
By (a),
I(2k)
n = 2k – r.
such that
I(2k – 1)
is true ⇒
∴ I(n – 1) is also true.
is true ⇒ I(2k – 2)
I(2k – r) = I(n)
is true ⇒ …⇒
x + x 2   x1 − x 2 
 x 1 + x 2 ,
sin x1 + sin x2 = 2 sin 1
 cos
 ≤ 2 sin

 2   2 
 2 
is true.
−π ≤ x1 − x 2 ≤ π .
∴ f(x) = sin x is convex on [0, π] . Last part follows from (i).
9.
,
R.H.S. =
∴
P(1) is true.
Let P(n) be the proposition : J1 − − "K + J − − SK + ⋯ + J' − "' − "K = J1 − + − " + ' − K
L.H.S. = 1 − − " = "
For P(1),
J1 − K = "
,
Assume P(k) is true for some k∈ N, that is,
J1 − − "K + J − − SK + ⋯ + J' − "' − "K = J1 − + − " + ' − K………(1)
J1 − − "K + J − − SK + ⋯ + J' − "' − "K + J − " − ""K
For P(k + 1),
= J1 − + − " + ' − K + J − " − ""K
= J1 − + − " + ' − K + J − × K − × !
= J1 − + − " + ' − K +
× − × !
9
= J1 − + − +
− +
−
K,
"
'
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true for all natural numbers n.
12.
Let Fn ( z) =
q
1− q
(1 − z ) +
q2
(1 − z )(1 − qz ) + ... +
1− q2
1− qn
(1 − z)(1 − qz)...(1 − q n −1z ) ,
1 + Fn(z) – Fn(qz) = (1 – qz)(1 – q2z) … (1 – qnz) .
Let P(n) be the proposition :
1 + F z
− F qz
= 1 +
For P(1),
qn
W
'W
1 − z
−
W
'W
1 − qz
= 1 − qz
∴ P(1)
is true.
Assume P(k) is true for some k∈ N, that is,
1 + F z
− F qz
= 1– qz
1–q z
…1–q z
………(1)
For P(k + 1),
F z
= F z
+
But,
WXYZ
'WXYZ
and F qz
= F qz
+
1 − z
1 − qz
… -1–q z.
WXYZ
'WXYZ
1 − qz
1 − qz
… -1–q z.
1 + F z
− F qz
= 1 + F z
− F qz
+
WXYZ
'WXYZ
1 − z
1 − qz
… -1–q z. −
= 1– qz
1–q z
…1–q z
+
WXYZ
'WXYZ
= 1– qz
1–q z
…1–q z
+
WXYZ
'WXYZ
1 − qz
1 − q z
… -1–q z.
1 − z
1 − qz
… -1–q z. −
WXYZ
'WXYZ
WXYZ
'WXYZ
1 − qz
1 − q z
… -1–q z., by (1)
1 − qz
… -1–q z.-1–q z. − -1–q z.
= -1– qz.-1–q z. …-1–q z. − q 1 − qz
… -1–q z.
= -1– qz.-1–q z. …-1–q z.-1–q z.
∴
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
14.
1+ =
[
[
[
[
=
[
+ ⋯+
,1 + +
[
Assume that 1 + +
[
Adding the term
1+ +
[
=
[
[\
[\
[\
[
[
+
[
[\
[\
[\^…]
[\^…]_
[
\
…]
[\^…]
[
\
[
\
…]
[
\
…]
+⋯+
=
+
=
[
\
…]
[\^…]
to both sides, we get:
[
\
…]
[\^…]_
=
[
\
…]
[\^…]
+
[
\
…]
[\^…]_
[
\
…]
_
[\^…]_
10
17.
1
1
1
1
 1

+ 3
+ ... +
+
+ ... +
=
3
2n + 1  2 − 2
( 2n ) − 2n  n + 1 n + 2
2n
n
Let P(n) be the proposition :
For P(1), L.H.S. =
1
2 ×1 + 1
+
1
2 −2
3
=
1
2
Assume P(k) is true for some k∈ N, that is,
1
=
∴ P(1)
= R.H.S.
1+1
is true.
1
1
1
1
 1

+ 3
+ ... +
+
+ ... +
=
3
2k + 1  2 − 2
( 2k ) − 2 k  k + 1 k + 2
2k
k
1
1
1 
 1
  1
+ 3
+ ... +
+
+ ... +
=0
−
3
2k + 1  2 − 2
( 2 k ) − 2k   k + 1 k + 2
2k 
k
or
k +1
1
1
1
1
1 
 1
  1
+
+ ... +
+
+
+ 3
+ ... +

−
3
2k + 3  2 − 2
[2( k + 1)] − 2( k + 1)   k + 2 k + 3
2k 2k + 1 2k + 2 
For P(k + 1),
=
=
=
=
k +1
2k + 3
k +1
2k + 3
(*)
−
−
k
2k + 1
k
2k + 1
1
( 2 k + 1)( 2 k + 3)
+
1
1
1 

− −
+
+
,
[ 2( k + 1)] − 2( k + 1)  k + 1 2 k + 1 2 k + 2 
+
+
1
by (*)
3
1
2( k + 1)( 2 k + 1)( 2 k + 3)
+
1
2( k + 1)( 2 k + 1)( 2 k + 3)
2( k + 1) + 1 − ( 2 k + 3)
2( k + 1)( 2k + 1)( 2 k + 3)
−
1
k +1
−
1
−
2k + 1
1
2( k + 1)
1
2( k + 1)( 2 k + 1)
∴ P(k+1) is true.
=0
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
18.
Let P(n) be the proposition :
a1 =
For P(1),
∴ P(1)
a+b
2
an = a +
2

= a + ( b − a ) 1 −

3
1 
2
1 


( b − a ) 1 − n  (1) , b n = a + ( b − a ) 1 +
 ( 2)


3
4 
3
2 × 4n 
2
1
a 1 + b (a + b ) / 2 + b a + 3b
2
1 

=
=
= a + ( b − a ) 1 +
, b 1 =


4
2
2
2
3
2×4
is true.
Assume P(k) is true for some k∈ N, i.e., a k = a +
1 
2
1 


( b − a )1 − k  (3) , b k = a + ( b − a )1 +
 ( 4)


3
4 
3
2 × 4k 
2
For P(k + 1),
a k + bk
a k +1 =
2
=
1
2
1
2
1
a + ( b − a )1 − k  + a + ( b − a )1 +



2
3
4 
3
2 × 4k
1
1
1
= a + ( b − a )  1 − k  +  1 +

3
4  
2 × 4k
b k +1 =
a k +1 + b k
2
=



, by (3) and (4)
  = a + 2 ( b − a ) 1 − 1 





3
4 k +1 
1
2
1
2
1
a + ( b − a ) 1 − k +1  + a + ( b − a )1 +



2
3
4 
3
2 × 4k
….(5)
 ,by (5) and (4)


1
1
2
2
1 
= a + ( b − a )  1 − k +1  + 1 + k +1  = a + ( b − a )1 +



3
4  
4 
3
2 × 4 k +1 
∴ P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
11
19.
1
pn-1 > pn > k > qn > qn-1 > 0 where p0 > k > q0, p n = ( p n −1 + q n −1 )
Let P(n) be the proposition :
For P(1),
2
Note that pC > ` > 0, qC =
a
a
;Z
>
a
;b
= qC
p = pC + qC > cpCq C = √k = k,k > 0 ,
q =
a
p > ` > 0, q =
Also,
pn q n = k 2 .
> 0.
;b
p = pC + qC < pC + pC = pC ,q =
and
;Z
a
;Z
<
a
=k
>0
∴ pC > p > ` > q > q C > 0 and P(1) is true.
Assume P(m) is true for some m∈ N, i.e., p1' > p1 > ` > q 1 > q 1' > 0 … 1
a
p1 = p1 + q1 < p1 + p1 = p1 ,q1 =
For P(m + 1),
;dYZ
>
p1 = p1 + q1 > cp1 q1 = √k = k,k > 0 ,
p1 > ` > 0, q 1 =
a
;dYZ
;d
= q1
q 1 =
Also,
a
a
;dYZ
<
a
=k
>0
∴ P(m + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
20.
(a)
u x
=
ff
f
…f'
!
∑;5 u x
= u; x + 1
− 1
Let S(p) be the statement :
For S(1), ∑5 u x
= u x
=
∴
f
!
u x + 1
− 1 =
= x,
f
!
−1=x
S(1) is true.
Assume S(k) is true for some k∈ N, i.e., ∑5 u x
= u x + 1
− 1 … 1
For S(k + 1),
∑
5 u x
= ∑5 u x
+ u x
= u x + 1
− 1 + u x
, by (1)
=
f
f
…f
−1+
=
f
f
…f
k + 1
+ x − 1
=
f
f
…f
f
!
!
!
= u x + 1
− 1
ff
f
…f
!
−1
∴
,
S(k + 1) is true.
By the Principle of Mathematical Induction, S(p) is true ∀ p ∈ N.
(b)
Obviously,
u JK =
Z
a
Z
a
Z
a
Z
a
J KgJ KhgJ Kh…gJ K'h
!
>0
12
Let P(n) be the proposition : u J K <
Z
a
J K
L. H. S. = u J K =
For P(1),
√
!
Assume P(k) is true for some k∈ N, i.e., u J K <
Z
a
For P(k + 1), u JK =
Z
a
gJ Kh
×
×n
<
√
<
√
Z
a
√
√
Z
a
≈ 0.5773502691896, ∴ P(1) is true.
… 1
Z
a
J KgJ KhgJ Kh…gJ K'hgJ Kh
Z
a
= ,R. H. S. =
=
√
√"a S"
o=
!
×
=
√
√
×
!=
=
Z
a
Z
a
Z
a
√√
×
Z
a
J KgJ KhgJ Kh…gJ K'h
!=
!
√
×n
×
Z
a
gJ Kh
√"a S
o
∴ P(k + 1) is true.
√
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
Since 0 < u; JK <
, hence −1 < u; JK − 1 <
c;
Now, from (a), put x = −
;
−1 < ∑5 u J− K <
−1 <
Z
a
J' K
!
−1 < −
+
Z
a
Z
a
Let P(n) be the proposition :
For P(1),
+⋯+
∙
∙
l1 = a < a + 1
and
Z
a
Z
a
∙∙8∙∙∙;'
∙"∙⋯;
∙∙8∙∙∙;'
∙"∙⋯;
Z
a
Z
a
;!
!<
>1−
and
c;
lk < a + 1
<
c;
−1
−1
c;
l n-1 < l n .
l1 = a < a + a = l 2
Assume P(k) is true for some k∈ N, i.e.,
For P(k + 1),
J' KgJ' KhgJ' Kh…gJ' K;'h
ln < a + 1
and
−1
+ ∙" + ∙"∙ + ⋯ +
1 > + ∙" + ∙"∙ + ⋯ +
21.
c;
J' KgJ' Kh
!
−1
∑;5 u J− K = u; J1 − K − 1 = u; J K − 1
,
c;
(1)
∴ P(1) is true.
and
l k-1 < l k
( 2)
l k +1 = a + l k < a + a + 1 < a + 2 a + 1 = a + 1
l k + 2 = a + l k +1 > a + l k = l k+1
∴ P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
n
22.
(a)
Prove of
∑i
i =1
2
=
1
6
n ( n + 1)( 2 n + 1) is omitted.
13
n
n
i =1
= ( n + 1)
(b)
n ( n + 1)
i =1
2
i =1
1
1
− n ( n + 1)( 2 n + 1) = n ( n + 1)( n + 2)
6
6
2
1
1+
Let P(n) be the proposition :
+
2
For P(1), L.H.S. = 1 ≤ 2 = R.H.S.
1
1+
2
1
+
3
+ ... +
1
1
k
+
3
n
1+
1
1
≤2 k +
2
k +1
1
= 2 k +1 − 2 k +1 + 2 k +
≤2 n
k +1
1
+
….(2)
3
=2
1
+ ... +
≤2 k
k
1
….(*)
, by (*)
k +1
= 2 k +1 −
k +1
( k + 1) − 2 k ( k + 1) + k
= 2 k +1 −
1
+ ... +
….(1)
∴ P(1) is true.
Assume P(k) is true for some k∈ N, i.e.,
For P(k + 1),
n
∑ i( n + 1 − i) = ( n + 1)∑ i − ∑ i
1⋅n + 2⋅(n – 1) + 3⋅( n – 2) + ⋅⋅⋅ + n⋅1 =
2( k + 1) − 2 k ( k + 1) − 1
k +1
(
k +1 −
k +1 − k
)
2
k +1
∴ P(k + 1) is true.
≥ 2 k +1
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N.
By (1),
1
6
( n + 1)( n + 2) =
1.n + 2( n − 1) + 3( n − 2) + ... + n.1
n
> [(1.n )( 2.( n − 1))...( n.1)]
1/ n
= [( n!)
]
2 1/ n
, by A.M.>G.M.
n
1
2
∴ n!<  n ( n + 1)( 2 n + 1)
6

2 n
By (2),
n
≥
1
1
1
1  
1
1
1 
+
+ ... +
×
× ... ×
1 +
 > 1 ×

n
2
3
n 
2
3
n
1/ n
1
= 
 n! 
1/ 2 n
n
∴
23.
n
  < n!
4
Let P(n) be the proposition :
For P(2),
L.H.S.=
√[Z √[a
√[Z √[a
+
=
+ ⋯+
R.H.S.,
∴
√[a c[q
Assume P(k) is true for some k∈ N\{1}, that is,
For P(k + 1),
√[Z √[a
+
=
=
√[a c[q
+⋯+
√[Z √[a
√[XsZ √[X
+
√[rsZ √[r
=
'
√[Z √[r
P(2) is true.
+
√[a c[q
√[X c[XYZ
+ ⋯+
=
''
t
+
-√[X 'c[XYZ .
't
'
=
√[XsZ √[X
√[Z √[X
-√[X 'c[XYZ .
'
√[Z '√[X +√[Z √[X √[Z '√[X √[X c[XYZ .-√[X 'c[XYZ .
'
√[Z '√[X where a3 is an arithmetic sequence.
+
=
'
√[Z √[X
√[X c[XYZ
'
√[Z '√[X [Z '[X
+
………(1)
, by (1)
-√[X 'c[XYZ .
[X '[XYZ
, where d is the common difference.
14
= − √a − √a − -√a − ca. = − -√a − ca .
t
t
t
= 't -√a − ca . = [
=
∴
Z '[XYZ
-√a − ca . = -
√[Z c[XYZ .-√[Z 'c[XYZ .
-√a − ca .
√[Z c[XYZ
P(k + 1) is true.
By the Principle of Mathematical Induction, P(n) is true ∀ n ∈ N\{1}.
15