Download APPENDIX B EXERCISES In Exercises 1–8, use the

APPENDIX B EXERCISES
In Exercises 1–8, use the principle of mathematical induction to establish the stated
identity involving the integer n.
1.
12 + 2 2 +  + n 2 =
n(n + 1)(2n + 1)
for n ≥ 1 .
6
2.
13 + 2 3 +  + n 3 =
n 2 (n + 1)2
for n ≥ 1 .
4
3.
2 + 2 2 +  + 2 n = 2 n +1 − 2 for n ≥ 1 .
4.
If r is a real number and r ≠ 1 , then 1 + r + r 2 +  + r n =
5.
1 + 5 + 9 +  + (4n + 1) = (n + 1)(2n + 1) for n ≥ 0 .
6.
1⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ 4 +  + n(n + 1)(n + 2) =
7.
1⋅1!+ 2 ⋅ 2!+ 3 ⋅ 3!+  + n ⋅ n! = (n + 1)!− 1 for n ≥ 1 .
8.
1 ⎞⎛
1⎞ ⎛
1 ⎞ n +1
⎛
for n ≥ 2 .
⎜⎝ 1 − 2 ⎟⎠ ⎜⎝ 1 − 2 ⎟⎠ ⎜⎝ 1 − 2 ⎟⎠ =
2
3
n
2n
r n +1 − 1
for n ≥ 0 .
r −1
n(n + 1)(n + 2)(n + 3)
for n ≥ 1 .
4
In Exercises 9–12, find a formula for the given expression and then prove that your
formula is correct using the principle of mathematical induction.
9.
1 − 2 + 3 − 4 +  + (2n − 1) − 2n
10.
1 − 4 + 9 − 16 +  + (2n − 1)2 − (2n)2
11.
1
1
1
1
+
+
++
1⋅ 2 2 ⋅ 3 3 ⋅ 4
n(n + 1)
12.
1⎞ ⎛
1⎞ ⎛
1⎞
⎛
⎜⎝ 1 − ⎟⎠ ⎜⎝ 1 − ⎟⎠ ⎜⎝ 1 − ⎟⎠
2
3
n
In Exercises 13–18, use the principle of mathematical induction to prove the stated
assertion involving the integer n.
13.
3n > 2 n +1 for n ≥ 1 .
14.
n! > 2 n for n ≥ 4 .
15.
If x is a real number, x > −1 and x ≠ 0 , then (1 + x)n > 1 + nx for n ≥ 2 .
16.
n 3 − n is a multiple of 3 for n ≥ 1 .
17.
4 n − 1 is a multiple of 3 for n ≥ 1 .
18.
10 2n + 2 ⋅10 2n −1 + 1 is a multiple of 11 for n ≥ 1 .
In Exercises 19 and 20, prove the stated property of exponents using the principle of
mathematical induction.
19.
For all real numbers a and b, (ab)n = a nb n for n ≥ 1 .
20.
For all real numbers a and b, (a m )n = a mn for all nonnegative integers m and n.
[Hint: Fix m and use induction on n ≥ 1 .]
21.
Using the principle of mathematical induction, prove that for any integer n ≥ 1
and any n real numbers x1 , x2 , …, xn , x1 +  + xn ≤ x1 +  + xn . [Note: The case
n = 2 is the Triangle Inequality for real numbers: x1 + x2 ≤ x1 + x2 .]
22.
If we glue three unit squares together as shown below, we get an L-tromino.
Consider a 2 n × 2 n checkerboard with one corner square removed. Using the principle of
mathematical induction, prove that for any integer n ≥ 1 , such a checkerboard can always
be completely covered by non-overlapping L-trominoes. (The L-trominoes can be rotated
if necessary.)
23.
Using the second principle of mathematical induction, prove that every positive
integer can be written as a product of an odd integer and a power of 2.
24.
Using the second principle of mathematical induction, prove that every positive
integer can be written as a sum of distinct powers of 2. [Hint: Every positive integer lies
between two consecutive powers of 2 (or is equal to one of these powers).]
25.
A chemist has an unlimited supply of 3 gram and 5 gram weights that she uses on
a 2-pan balance scale. Using the second principle of mathematical induction, prove that
she can accurately weigh any amount greater than or equal to 8 grams. [Hint:
k + 1 = (k − 2) + 3 ]
26.
A unit fraction is the reciprocal 1 n of a positive integer n. An Egyptian fraction
is any rational number that can be written as a sum of distinct unit fractions. For example,
2 1 1
= +
3 2 6
and
4 1 1 1
= + +
5 2 4 20
are Egyptian fractions.
Prove that every rational number between 0 and 1 is an Egyptian fraction. [Hint:
If 0 < n m < 1 , let 1 d be the largest unit fraction such that 1 d ≤ n m . Use the second
principle of mathematical induction on the numerator and consider n m − 1 d for the
induction step.]
27.
A sequence x1 , x2 , …, xn of real numbers satisfies x1 = 3 , x2 = 5 , and
xn +1 = 3xn − 2xn −1 for n ≥ 2 . Using the second principle of mathematical induction, prove
that xn = 2 n + 1 for all integers n ≥ 1 .
28.
A sequence x1 , x2 , …, xn of real numbers satisfies x1 = 2 , x2 = 3 , and
xn +1 = xn xn −1 for n ≥ 2 . Using the second principle of mathematical induction, prove
that xn + 6 = xn for all integers n ≥ 1 .
In Exercises 29–32, find the flaw in the induction “proof”.
29.
Let’s prove that n + 1 < n , where n is a positive integer. For n = k , we assume
that k + 1 < k . Then for n = k + 1 we have (k + 1) + 1 < k + 1 . So the result follows by the
principle of mathematical induction.
30.
All real numbers are equal. More specifically, in any set of n real numbers, all
numbers in the set are equal. Clearly if we have only one real number, it is equal to itself.
Now assume that in any set of k real numbers, all numbers in the set are equal. Consider a
set {x1 , x2 , …, xk , xk +1 } of k + 1 real numbers. By the induction hypothesis,
x1 = x2 =  = xk and also x2 =  = xk = xk +1 . Hence x1 = x2 =  = xk = xk +1 and so all of
the numbers are equal. The result follows by the principle of mathematical induction.
31.
For all positive integers n, if the maximum of the positive integers a and b is n,
then a = b . Let max(a, b) denote the maximum of a and b . If max(a, b) = 1 , then
a = b = 1 . Assume that if max(a, b) = k , then a = b . Now let max(a, b) = k + 1 . Then
max(a − 1, b − 1) = k and so a − 1 = b − 1 by the induction hypothesis. Therefore, a = b .
This completes the induction step and so the result is true for all integers n ≥ 1 .
32.
If a is a nonzero real number, then a n = 1 for all n ≥ 0 . Certainly a 0 = 1 . Now
assume that a k = 1 for all k ≤ n . Then we may write n + 1 = p + q for p, q ≤ n . Hence,
a n +1 = a p + q = a p a q = 1⋅1 = 1 and so the result follows by the second principle of
mathematical induction.
33.
Professor Snarf’s linear algebra class meets five days a week, Monday through
Friday. One Friday, he said to the class, “There will be a surprise quiz next week. You
won’t know which day it will occur until the actual day of the quiz.”
Annie Zog, one of the students in the class, whispered to Bert, the student sitting
beside her, “Ha! That means we can’t have a quiz at all next week!”
“What are you talking about?” replied Bert, perplexed.
“Induction!” Annie announced. “I can prove by induction that Snarfie can’t give
us a quiz next week.”
“You’re kidding,” said Bert. “Convince me!”
Annie continued, “Number the days next week like this: Monday=1, Tuesday=2,
…, Friday=5. Let S(n) be the statement ‘There can’t be a quiz on day 5 − n .’ Now S(0)
says ‘There can’t be a quiz on day 5.’ That’s Friday and if we haven’t had the quiz by
Thursday, then we’ll know it’s on Friday and it won’t be a surprise. So S(0) is true.
Next, assume that S(k) is true for 0 ≤ k ≤ 3 . That is, there can’t be a surprise quiz on
days 5 − k through 5. But then day 5 − k − 1 is the latest possible day for the quiz and, if
we haven’t already had the quiz by then, we’ll know it’s coming and it won’t be a
surprise. So there can’t be a quiz on day 5 − k − 1 = 5 − (k + 1) either. That’s the
induction step and so we can’t have a quiz any day next week!”
“Beautiful!” exclaimed Bert. “You’ve caught Snarf at his own game! Now we
don’t have to study for a quiz.”
It certainly came as a surprise to Annie and Bert when the quiz happened on
Wednesday! Can you find a flaw in Annie’s reasoning? Is there a flaw?