SOLUTION OF HW3
MINGFENG ZHAO
September 25, 2011
1. [Page 36, Problem 5] Guess a general law which simplifies the product
1
1
1
1
1−
1−
··· 1 − 2 .
1−
4
9
16
n
and prove it by induction.
Proof. When n = 2, we know
1−
1
1
3
=1− = .
22
4
4
When n = 3, we have
1
1−
4
1
3
1
3 8
2
4
1− 2 = · 1−
= · = = .
3
4
9
4 9
3
6
When n = 4, we have
1
5
1
1
2
2 15
1
= .
1−
1−
1− 2 = · 1−
= ·
4
9
4
3
16
3 16
8
When n = 5, we have
1−
1
4
1
1
1
1
5 24
3
6
5
1−
1−
1− 2 = · 1−
= ·
= =
.
9
16
5
8
25
8 25
5
10
We guess the general law is:
(1)
1
1−
4
1
1
1
n+1
1−
1−
··· 1 − 2 =
,
9
16
n
2n
for all n ≥ 2.
Now we prove (??) by induction. When n = 2, we have known that
1−
1
3
2+1
= =
.
22
4
2·2
1
2
MINGFENG ZHAO
Now we assume that for n = k, and k ≥ 2, we have
1−
(2)
1
4
k+1
1
1
1
1−
1−
··· 1 − 2 =
.
9
16
k
2k
For n = k + 1, we have
1
1−
1−
9
1
1
=
1−
1−
1−
4
9
k+1
1
=
· 1−
2k
(k + 1)2
1−
1
4
=
k + 1 (k + 1)2 − 1
·
2k
(k + 1)2
=
k + 1 k 2 + 2k + 1 − 1
·
2k
(k + 1)2
=
k + 1 k 2 + 2k
·
2k
(k + 1)2
=
k + 1 k(k + 2)
·
2k
(k + 1)2
=
k+2
2(k + 1)
=
(k + 1) + 1
.
2(k + 1)
1
(k + 1)2
1
1
1
1−
··· 1 − 2
16
k
(k + 1)2
1
16
··· 1 −
By the assumption ?? for n = k
Therefore, by induction, we can conclude that
1
1
1
1
n+1
1−
1−
1−
··· 1 − 2 =
,
4
9
16
n
2n
for all n ≥ 2.
2. [Page 36, Problem 6] Let A(n) denote the statement : 1 + 2 + · · · + n = 81 (2n + 1)2 .
a. Prove that if A(k) is true for an integer k, then A(k + 1) is also true.
b. Criticize the statement: “ By induction it follows that A(n) is true for all n ”.
c. Amend A(n) by changing the equality to an inequality that is true for all positive integers n.
SOLUTION OF HW3
3
Proof. a. Assume A(k) is true for an integer k, that is, we have
1 + 2 + ··· + k =
(3)
1
(2k + 1)2 .
8
Let n = k + 1, we know that
1 + 2 + · · · + (k + 1)
=
1 + 2 + · · · + k + (k + 1)
=
1
(2k + 1)2 + (k + 1)
8
=
1
[(2k + 1)2 + 8(k + 1)]
8
=
1 2
[4k + 4k + 1 + 8k + 8]
8
=
1 2
[4k + 12k + 9]
8
=
1
(2k + 3)2
8
=
1
[(2(k + 1) + 1]2 .
8
By (??)
That is, A(k + 1) is also true.
b. The the statement: “ By induction it follows that A(n) is true for all n ” is wrong. Because we
know that when n = 1, we have
1<
1
9
(2 · 1 + 1)2 = .
8
8
c. Claim: For all n ≥ 1, we have
1 + 2 + ··· + n ≤
1
(2n + 1)2 .
8
We prove this Claim by induction. Firstly, look at n = 1, in the proof of the part b, we have known
that
1<
1
9
(2 · 1 + 1)2 = .
8
8
4
MINGFENG ZHAO
That is, the Claim is true for n = 1. We assume that for n = k, the Claim is also true, that is,
1 + 2 + ··· + k ≤
(4)
1
(2k + 1)2 .
8
Now look at n = k + 1, we know that
1 + 2 + · · · + (k + 1)
=
1 + 2 + · · · + k + (k + 1)
≤
1
(2k + 1)2 + (k + 1)
8
=
1
[(2k + 1)2 + 8(k + 1)]
8
=
1 2
[4k + 4k + 1 + 8k + 8]
8
=
1 2
[4k + 12k + 9]
8
=
1
(2k + 3)2
8
=
1
[(2(k + 1) + 1]2 .
8
By (??)
That is, the Claim is true for n = k + 1. Therefore, by induction, we can conclude that for all
n ≥ 1, we have
1 + 2 + ··· + n ≤
1
(2n + 1)2 .
8
3. [Page 40, Problem 6]
Pn
k=1
k2 =
n3
3
+
n2
2
+ n6 .
Proof. Firstly, we notice that for all k ≥ 1, we have
k 3 − (k − 1)3
= k 3 − [k 3 − 3k 2 + 3k − 1]
=
3k 2 − 3k + 1.
Then we get
k2 =
1
1 3
1
[k − (k − 1)3 ] + 3k − 1 = [k 3 − (k − 1)3 ] + k − .
3
3
3
SOLUTION OF HW3
5
Hence, for all n ≥ 1, we get
n
X
k2
=
k=1
n X
1
3
k=1
=
n
X
1
k=1
=
3
[k 3 − (k − 1)3 ] + k −
[k 3 − (k − 1)3 ] +
n
X
1
3
k−
k=1
n
X
1
k=1
3
By the results of the Exercise 2, in Page 40
2
n
1X 3
n
n
n
[k − (k − 1)3 ] +
+
−
3
2
2
3
k=1
By the results of the Exercise 3 and Exercise 4, in Page 40
=
2
1
n
n
n
3
3
· [n − (1 − 1) ] +
+
−
3
2
2
3
=
1 3 n2
n n
·n +
+ −
3
2
2
3
=
n3
n2
n
+
+ .
3
2
6
By the result of the Exercise 2, in Page 40
Therefore, we conclude that for all n ≥ 1, we have
n
X
k2 =
k=1
n3
n2
n
+
+ .
3
2
6
4. [Page 41, Problem 12] Guess and prove a general rule which simplifies the sum
n
X
k=1
1
.
k(k + 1)
Proof. When n = 1, we have
1
X
k=1
1
1
1
1
=
= =
.
k(k + 1)
1(1 + 1)
2
1+1
When n = 2, we have
2
X
k=1
1
1
1
1 1
4
2
2
=
+
= + = = =
.
k(k + 1)
1(1 + 1) 2(2 + 1)
2 6
6
3
2+1
6
MINGFENG ZHAO
When n = 3, we have
3
X
k=1
2
X
1
1
2
1
9
3
3
1
=
+
= +
=
= =
.
k(k + 1)
k(k + 1) 3(3 + 1)
3 12
12
4
3+1
k=1
So we guess the general law is:
n
X
k=1
n
1
=
.
k(k + 1)
n+1
In fact, for all k ≥ 1, we know that
1
(k + 1) − k
1
1
−
=
=
.
k k+1
k(k + 1)
k(k + 1)
So for all n ≥ 1, we get
n
X
k=1
1
k(k + 1)
=
n X
1
k=1
=
=
1
−
k k+1
n X
1
1
−
−
By the result of the Exercise 2, in Page 40
k+1 k
k=1
1
−1
By the result of the Exercise 2, in Page 40
−
n+1
1
n+1
=
1−
=
n
.
n+1
Therefore, we conclude that for all n ≥ 1, we have
n
X
k=1
1
n
=
.
k(k + 1)
n+1
5. [Page 57, Problem 12 b] Show that
Pm
k=0
1−xn+1
1−x ,
x 6= 1, is a polynomial by converting it to the form
ak xk for a suitable m.
Proof. Claim: For all n ≥ 1, we have
n
X
1 − xn+1
= 1 + x + x2 + · · · + xn =
xi .
1−x
i=0
SOLUTION OF HW3
7
Here we use x0 = 1. When n = 1, we know that
1 − x1+1
1 − x2
(1 − x)(1 + x)
=
=
= 1 + x.
1−x
1−x
1−x
So the Claim is true for n = 1. We assume that when n = k, the Claim is true, that is,
k
X
1 − xk+1
= 1 + x + x2 + · · · + xk =
xj .
1−x
j=0
(5)
Now look at n = k + 1, we know that
1 − x(k+1)+1
1−x
=
1 − x · xk+1
1−x
=
1 + x · [−xk+1 ]
1−x
=
1 + x · [1 − xk+1 − 1]
1−x
=
1 + x[1 − xk+1 ] − x
1−x
=
(1 − x) + x[1 − xk+1 ]
1−x
=
1+x·
=
1+x·
1 − xk+1
1−x
k
X
xj
j=0
=
1+
k
X
xj+1
By the result of the Exercise 2, in Page 40
j=0
=
1+
k+1
X
xi
i=1
=
k+1
X
xi .
i=0
That is, the Claim is true for n = k + 1. Therefore, by induction, we can conclude that for all
n ≥ 1, we have
1 − xn+1
= 1 + x + x2 + · · · + xn .
1−x
8
MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu