10-4 Mathematical Induction
Use mathematical induction to prove that each conjecture is true for all positive integers n.
1. 3 + 5 + 7 + . . . + (2n + 1) = n(n + 2)
SOLUTION: Let Pn be the statement 3 + 5 + 7 + . . . + (2n + 1) = n(n + 2). Because 3 = 1(1 + 2) is a true statement, Pn is true for
n = 1. Assume that Pk : 3 + 5 + 7 + . . . + (2k + 1) = k(k + 2) is true for a positive integer k. Show that Pk + 1 must be
true.
Add [2(k + 1) + 1] to both sides of the equation. Simplify the right side. Then manipulate the side to get terms
replacing k with k + 1 in the original expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 3 + 5 + 7 + . . . + (2n + 1) = n(n + 2) is
true for all positive integers n.
3. 2 + 22 + 23 +. . . + 2n = 2(2n – 1)
SOLUTION: 2
3
n
n
1
Let Pn be the statement 2 + 2 + 2 + . . . + 2 = 2(2 – 1). Because 2 = 2(2 – 1) is a true statement, Pn is true for n
2
3
k
k
= 1. Assume that 2 + 2 + 2 + . . . + 2 = 2(2 – 1) is true for a positive integer k. Show that Pk + 1 must be true.
Add
to both sides of the equation. Simplify the right side. Then manipulate the side to get terms replacing k with
k + 1 in the original expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
2
3
n
n
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 2 + 2 + 2 + . . .+ 2 = 2(2 – 1) is
true for
all positive
n.
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5. 1 + 4 + 7 + … + (3n – 2) =
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 3 + 5 + 7 + . . . + (2n + 1) = n(n + 2) is
10-4
Induction
trueMathematical
for all positive integers
n.
3. 2 + 22 + 23 +. . . + 2n = 2(2n – 1)
SOLUTION: 2
3
n
n
1
Let Pn be the statement 2 + 2 + 2 + . . . + 2 = 2(2 – 1). Because 2 = 2(2 – 1) is a true statement, Pn is true for n
2
3
k
k
= 1. Assume that 2 + 2 + 2 + . . . + 2 = 2(2 – 1) is true for a positive integer k. Show that Pk + 1 must be true.
Add
to both sides of the equation. Simplify the right side. Then manipulate the side to get terms replacing k with
k + 1 in the original expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
2
3
n
n
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 2 + 2 + 2 + . . .+ 2 = 2(2 – 1) is
true for all positive integers n.
5. 1 + 4 + 7 + … + (3n – 2) =
SOLUTION: Let Pn be the statement 1 + 4 + 7 + . . . + (3n – 2) =
Pn is true for n = 1. Assume that 1 + 4 + 7 + . . . + (3k – 2) =
. Because
is a true statement, is true for a positive integer k. Show that
Pk + 1 must be true.
Add [3(k + 1) – 2] to both sides of the equation. Simplify the right side. Then manipulate the side to get terms
replacing k with k + 1 in the original expression.
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This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
2
for Mathematical
n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 2 + 2
10-4
Induction
3
n
n
+ 2 + . . .+ 2 = 2(2 – 1) is
true for all positive integers n.
5. 1 + 4 + 7 + … + (3n – 2) =
SOLUTION: Let Pn be the statement 1 + 4 + 7 + . . . + (3n – 2) =
. Because
Pn is true for n = 1. Assume that 1 + 4 + 7 + . . . + (3k – 2) =
is a true statement, is true for a positive integer k. Show that
Pk + 1 must be true.
Add [3(k + 1) – 2] to both sides of the equation. Simplify the right side. Then manipulate the side to get terms
replacing k with k + 1 in the original expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 1 + 4 + 7 + . . . + (3n – 2) =
is true for all positive integers n.
7. 1 + 2 + 4 + … + 2n – 1 = 2n – 1
SOLUTION: n –1
Let Pn be the statement 1 + 2 + 4 + … + 2
k –1
Assume that 1 + 2 + 4 + … + 2
n
1
= 2 – 1. Because 1 = 2 – 1 is a true statement, Pn is true for n = 1.
k
= 2 – 1 is true for a positive integer k. Show that Pk + 1 must be true.
Add
to both sides of the equation. Simplify the right side. Then manipulate the side to get terms replacing k
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with kManual
+ 1 in- Powered
the original
expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 1 + 4 + 7 + . . . + (3n – 2) =
10-4 Mathematical
Induction
is true for all positive integers n.
7. 1 + 2 + 4 + … + 2n – 1 = 2n – 1
SOLUTION: n –1
Let Pn be the statement 1 + 2 + 4 + … + 2
k –1
Assume that 1 + 2 + 4 + … + 2
n
1
= 2 – 1. Because 1 = 2 – 1 is a true statement, Pn is true for n = 1.
k
= 2 – 1 is true for a positive integer k. Show that Pk + 1 must be true.
Add
to both sides of the equation. Simplify the right side. Then manipulate the side to get terms replacing k
with k + 1 in the original expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
n –1
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 1 + 2 + 4 + … + 2
for all positive integers n.
9. +
+ +. . .+ n
= 2 – 1 is true
= 1 –
SOLUTION: Let Pn be the statement
for n = 1. Assume that
+ + + + + . . . +
= 1 –
. Because
= 1 – is a true statement, Pn is true
+ . . . +
= 1 –
is true for a positive integer k. Show that Pk + 1 must
be true.
Add
to both sides of the equation. Simplify the right side. Then manipulate the side to get terms replacing k with
k + 1 in the original expression.
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This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
n –1
for Mathematical
n = 2, n = 3, and so Induction
on. That is, by the principle of mathematical induction, 1 + 2 + 4 + … + 2
10-4
n
= 2 – 1 is true
for all positive integers n.
9. +
+ +. . .+ = 1 –
SOLUTION: Let Pn be the statement
for n = 1. Assume that
+ + + + + . . . +
= 1 –
. Because
= 1 – is a true statement, Pn is true
+ . . . +
= 1 –
is true for a positive integer k. Show that Pk + 1 must
be true.
Add
to both sides of the equation. Simplify the right side. Then manipulate the side to get terms replacing k with
k + 1 in the original expression.
This final statement is exactly Pk + 1, so Pk + 1 is true. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true
for n = 2, n = 3, and so on. That is, by the principle of mathematical induction,
+ + + . . . +
= 1 –
is true for all positive integers n.
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