MTH 181
Section: 10 AM or Noon
(1pt) Name _____Solution Key _________
WS 04: Mathematical Induction (Sections 4.1–4.2) Version 3
Prove each of the formulas below by mathematical induction. To receive full credit, you must clearly show
1. The Basis Step
2. The Induction Step:
(a) The Induction Hypothesis written out in full!
(b) The Induction Step Proof. Write out in full the statement that must be proved in this step and be sure
to clearly show where the induction hypothesis is used within your proof for this step!
1. (8 pt) Prove ! 71 + 7 2 + 7 3 + ...+ 7 n =
7 n+1 − 7
for all ! n ≥ 1 .
6
?
(
)
Solution: Base Step: Show ! P(1) is true. Now, since ! 7 = 71 = 71+1 − 7 6 = 42 6 = 7 , ! P(1) is true!
Induction Hypothesis: Assume ! P(n) is true for some ! n = k where ! k ≥ 1 ; that is,
(
)
assume ! P(k) : 71 + 7 2 +!+ 7 k = 7 k+1 − 7 6 is true where ! k ≥ 1 .
Induction Step Proof: Use the induction hypothesis to prove ! P(n) is true for ! n = k + 1 where ! k ≥ 1 ; that is,
(
)
prove ! P(k + 1) : 71 + 7 2 +!+ 7 k+1 = 7 k+2 − 7 5 is true where ! k ≥ 1
Now,
7 + 7 +!+ 7
1
2
k+1
⎛
⎞
⎛ 7 k+1 − 7 ⎞
1
2
k
k+1
= ⎜ 7!#########
+ 7 +!+ 7" ⎟ + 7 = ⎜
+ 7 k+1 (by the Induction Hypothesis)
⎟
⎝ 6 ⎠
⎜⎝
⎟⎠
= ( 7 k+1 −7 ) 6
1⋅ 7 k+1 − 7 6 ⋅ 7 k+1 7 ⋅ 6 k+1 − 7 7 k+2 − 7
+
=
=
(by Rules of Algebra)
6
6
6
6
Thus, ! P(k) ⇒ P(k + 1) . Therefore, by the Principle of Mathematical Induction, ! P(n) is true for all integers ! n ≥ 1 . ! !
=
2. (8 pt) Prove !
1
1
1
1
n
+
+
+ +
=
for all ! n ≥ 1 .
1⋅ 7 7 ⋅13 13⋅19
( 6n − 5 )( 6n + 1) 6n + 1
L
?
Solution: Base Step: Show ! P(1) is true. Now, since ! 1 7 = 1 (1⋅ 7 ) = 1 ( 6 ⋅1+ 1) = 1 7 , ! P(1) is true!
Induction Hypothesis: Assume ! P(n) is true for some ! n = k where ! k ≥ 1 ; that is,
assume ! 1 (1⋅ 7 ) + 1 ( 7 ⋅13) +!+ 1 ⎡⎣( 6k − 5 ) ( 6k + 1) ⎤⎦ = k ( 6k + 1) is true where ! k ≥ 1 .
Induction Step Proof: Use the induction hypothesis to prove ! P(n) is true for ! n = k + 1 where ! k ≥ 1 ; that is,
prove ! P(k + 1) : 1 (1⋅ 7 ) + 1 ( 7 ⋅13) +!+ 1 ⎡⎣( 6(k + 1) − 5 ) ( 6(k + 1) + 1) ⎤⎦ =
k +1
k +1
=
is true where ! k ≥ 1 .
6(k + 1) + 1 6k + 7
⎛
⎞
⎜ 1
⎟
1
1
1
1
1
1
+
+!+
=⎜
+
+!+
⎟+
Now, 1⋅ 7 7 ⋅13
( 6(k + 1) − 5 )( 6(k + 1) + 1) ⎜ 1⋅ 7 7 ⋅13
( 6k − 5 )( 6k + 1) ( 6k + 1)( 6k + 7 )
!######################" ⎟
⎝
⎠
= k ( 6 k+1)
1
⎛ k ⎞
=⎜
+
(by the Induction Hypothesis)
⎝ 6k + 1 ⎟⎠ ( 6k + 1) ( 6k + 7 )
=
( 6k + 1) ( k + 1)
k ( 6k + 7 )
1
6k 2 + 7k + 1
+
=
=
( 6k + 1)( 6k + 7 ) ( 6k + 1)( 6k + 7 ) ( 6k + 1)( 6k + 7 ) ( 6k + 1) ( 6k + 7 )
=
k +1
(by Rules of Algebra)
6k + 7
Thus, ! P(k) ⇒ P(k + 1) . Therefore, by the Principle of Mathematical Induction, ! P(n) is true for all integers ! n ≥ 1 . ! !
1
MTH 181
Section: 10 AM or Noon
(1pt) Name _____Solution Key _________
WS 04: Mathematical Induction (Sections 4.1–4.2) Version 3
3. (8 pt) Prove by mathematical induction that ! 2n + 1 < 2 n for all ! n ≥ 3 .
Solution:
?
Basis Step: Show ! P(3) is true. Now, since ! 7 = 2(3) + 1 < 2 3 = 8 , ! P(3) true!
Induction Hypothesis: Assume ! P(n) is true for some ! n = k where ! k ≥ 1 ; that is,
assume ! 2k + 1 < 2 k is true where ! k ≥ 3 is true.
Induction Step Proof: Use the induction hypothesis to prove ! P(n) is true for ! n = k + 1 where ! k ≥ 3 ; that is,
prove ! P(k + 1) : 2(k + 1) + 1 < 2 k+1 is true where ! k ≥ 3
⎛
⎞
k
Now, 2(k + 1) + 1 = 2k + 2 + 1 = 2k + 1+ 2 = ⎜ 2k
+
1
!##k" ⎟ + 2 < 2 + 2 (by the Induction Hypothesis)
⎝ <2 ⎠
< 2 k + 2 k (since 2 < 2 k when k ≥ 3)
< 2 ⋅ 2 k = 2 k+1 (by Rules of Algebra)
Thus, ! P(k) ⇒ P(k + 1) . Hence, by the Principle of Mathematical Induction, ! P(n) is true for all integers ! n ≥ 3 .
!!
Extra Credit (25 pt): Do the following problems for extra credit (added to your quiz-worksheet scores).
1. Prove by mathematical induction that ! 2n 3 + 3n 2 + n is divisible by 6 for all integers ! n ≥ 1 .
No credit if done by another method!
Solution: Base Step: Show ! P(1) is true. Now, ! 2 (1) + 3(1) + 1 = 6 which is divisible by 6!
Induction Hypothesis: Assume ! P(n) is true for some ! n = some k ≥ 1 ; that is,
3
2
assume ! P(k) : 2k 3 + 3k 2 + k is divisible by 6 so that ! 2k 3 + 3k 2 + k = 6s for some s ∈! is true.
Induction Step Proof: Use the induction hypothesis to prove ! P(n) is true for ! n = k + 1 where ! k ≥ 1 ; that is,
prove ! P(k + 1) : 2(k + 1)3 + 3( k + 1) + ( k + 1) is divisible by 6 is true when ! k ≥ 1 .
2
Proof:
2
2(k + 1)31 + 3( k + 1) + ( k + 1) = 2 k 3 + 3k 2 + 3k + 1 + 3 k 2 + 2k + 1 + ( k + 1)
(
= ( 2k 3 + 6k 2
) (
+ 6k + 2 ) + ( 3k
2
)
+ 6k + 3) + ( k + 1)
⎛ 3
⎞
2
2
= ⎜ 2k
+
3k
+
k
!#
#"##
$ ⎟ + ( 6k + 6k + 6 ) by collecting like terms and regrouping
⎝
⎠
= 6s
!
= 6s + 6 ( k 2 + k + 1) by the induction hypothesis and factoring
= 6 ( s + k 2 + k + 1) by factoring
= 6t where t = s + k 2 + k + 1 ∈! since s, k, 1 are integers.
Thus, ! 2(k + 1)31 + 3( k + 1) + ( k + 1) = 6t, t ∈! .
Therefore, ! P(k) ⇒ P(k + 1) is true! Hence, by PMI, ! P ( n ) is true for all positive integer n.! !
2
2
MTH 181
Section: 10 AM or Noon
(1pt) Name _____Solution Key _________
WS 04: Mathematical Induction (Sections 4.1–4.2) Version 3
2. Prove by mathematical induction that ! 1+ 3n ≤ 4 n for every integer ! n ≥ 0 .
No credit if proved by another method!
Solution: Base Step: Show ! P(0) is true. Now, ! 1+ 3( 0 ) = 1 ≤ 1 = 4 0 . So ! P(0) is true!
Induction Hypothesis: Assume ! P(n) is true for ! n = some k ≥ 0 ; that is,
assume ! P(k) : 1+ 3k ≤ 4 k where ! k ≥ 0 is true.
Induction Step Proof: Use the induction hypothesis to prove ! P(n) is true for ! n = k + 1 where ! k ≥ 0 ; that is,
prove ! P(k + 1) : 1+ 3( k + 1) < 2 k+1 where ! k ≥ 0 .
Proof:
k
k
! 1+ 3( k + 1) = 1+
3k
!k ##
" + 3 ≤ 4 + 3 = 1⋅ 4 +
≤4 by the
ind. hyp.
3⋅1
!"
≤ 1⋅ 4 k + 3⋅ 4 k = (1+ 3) 4 k = 4 ⋅ 4 k = 4 k+1
Since
k≥0, 1≤4 k
Therefore, ! P(k) ⇒ P(k + 1) is true! Hence, by PMI, ! P ( n ) is true for all integers ! n ≥ 0 .! !
1⎞ ⎛
1⎞ ⎛
1⎞ ⎛
1 ⎞ n +1
⎛
3. Prove by mathematical induction that ! ⎜ 1− 2 ⎟ ⋅ ⎜ 1− 2 ⎟ ⋅ ⎜ 1− 2 ⎟ !⎜ 1− 2 ⎟ =
for all integers ! n ≥ 2 .
⎝ 2 ⎠ ⎝
⎠
⎝
⎠
⎝
3
4
n ⎠
2n
No credit if proved by another method!
3
1
1 ? 2 +1 3
= 1− = 1− 2 =
= . So ! P(2) is true!
4
4
2 2⋅2 4
Induction Hypothesis: Assume ! P(n) is true for ! n = some k ≥ 2 ; that is,
Solution: Base Step: Show ! P(2) is true. Now, !
1⎞ ⎛
1⎞ ⎛
1⎞ ⎛
1 ⎞ k +1
⎛
assume ! ⎜ 1− 2 ⎟ ⋅ ⎜ 1− 2 ⎟ ⋅ ⎜ 1− 2 ⎟ !⎜ 1− 2 ⎟ =
where ! k ≥ 2 is true.
⎝ 2 ⎠ ⎝
3 ⎠ ⎝
4 ⎠ ⎝
k ⎠
2k
Induction Step Proof: Use the induction hypothesis to prove ! P(n) is true for ! n = k + 1 where ! k ≥ 2 ; that is,
1⎞ ⎛
1⎞ ⎛
1⎞ ⎛
1⎞ ⎛
1 ⎞ ( k + 1) + 1
k+2
⎛
prove ! ⎜ 1− 2 ⎟ ⋅ ⎜ 1− 2 ⎟ ⋅ ⎜ 1− 2 ⎟ !⎜ 1− 2 ⎟ ⋅ ⎜ 1−
=
where ! k ≥ 2 .
2⎟ =
⎝ 2 ⎠ ⎝
3 ⎠ ⎝
4 ⎠ ⎝
k ⎠ ⎝ ( k + 1) ⎠
2 ( k + 1) 2 ( k + 1)
Proof:
1⎞ ⎛
1⎞ ⎛
1⎞ ⎛
1⎞ ⎛
1 ⎞ ⎛ 1+ k ⎞ ⎛
1 ⎞
⎛
=
⋅
1−
by the Induction Hypothesis
⎜⎝ 1− 2 ⎟⎠ ⋅ ⎜⎝ 1− 2 ⎟⎠ ⋅ ⎜⎝ 1− 2 ⎟⎠ !⎜⎝ 1− 2 ⎟⎠ ⋅ ⎜ 1−
⎜
⎟
2⎟
⎝ 2k ⎠ ⎜⎝ ( k + 1)2 ⎟⎠
2
3
4
k
k
+
1
(
)
⎝
⎠
!##########################"
=1+k
2k
2
⎛ 1+ k ⎞ ( k + 1) − 1 ⎛ 1+ k ⎞ k + 2k
=⎜
⋅
=
⋅
by Rules of Algebra
⎜⎝
⎟
⎝ 2k ⎟⎠ ( k + 1)2
2k ⎠ ( k + 1)2
2
!
(1+ k ) k ( k + 2 ) k + 2
=
by Rules of Algebra
2 k ( k + 1) ( k + 1) 2 ( k + 1)
PMI, ! P ( n ) is true for all integers ! n ≥ 2 .! !
=
Therefore, ! P(k) ⇒ P(k + 1) is true! Hence, by
3