Induction Conclusion
Consider
n
=
k
+ 1. Then
1

3 + 5

7 +
· · ·
+ (

1)
k

1
(2
k

1) + (

1)
k
(2(
k
+ 1)

1)
=
(

1)
k

1
k
+ (

1)
k
(2
k
+ 1)
(by Induction Hypothesis)
=
(

1)
k
(

k
) + (

1)
k
(2
k
+ 1)
=
(

1)
k
(

k
+ 2
k
+ 1)
=
(

1)
k
(
k
+ 1)
.
Therefore the result holds for
n
=
k
+ 1
.
Therefore, by the Principle of Mathematical Induction, 1

3 + 5

7 +
· · ·
+ (

1)
n

1
(2
n

1)
=
(

1)
n

1
n
for
all
n
∈
P
.
Problem 460:
Prove that
gcd(
f
n
, f
n
+1
) = 1 for all
n
∈
P
.
Solution:
We shall prove the result by induction on
n
.
Base Case
When
n
= 1, gcd(
f
1
, f
2
) = gcd(1
,
1) = 1, so the result holds.
Induction Hypothesis
Assume that gcd(
f
k
, f
k
+1
) = 1 for some
k
∈
P
,
k
≥
1.
Induction Conclusion
Consider
n
=
k
+ 1.
We want to show that gcd(
f
k
+1
, f
k
+2
) = 1.
Let
d
= gcd(
f
k
+1
, f
k
+2
), so
d

f
k
+1
and
d

f
k
+2
.
Now
f
k
+2
=
f
k
+
f
k
+1
, so
f
k
=
f
k
+2

f
k
+1
.
Hence
d

f
k
by Prop. 2.11.ii.
Since
d

f
k
and
d

f
k
+1
, it follows from the Induction Hypothesis that
d
= 1, since gcd(
f
k
, f
k
+1
) = 1.
(Alternatively, we could have said that
gcd(
f
k
+1
, f
k
+2
) = gcd(
f
k
+1
, f
k
+1
+
f
k
) = gcd(
f
k
+1
,
1(
f
k
+1
) +
f
k
) = gcd(
f
k
+1
, f
k
)
by Proposition 2.21, and then used the Induction Hypothesis.)
Therefore, the result is true for
n
=
k
+ 1.
Therefore, by the Principle of Mathematical Induction, gcd(
f
n
, f
n
+1
) = 1 for all
n
∈
P
.
Problem 467:
Prove that
f
n
+5
≡
3
f
n
(mod 5)
for all
n
∈
P
Solution 1:
Using the recurrence relation for the Fibonacci sequence repeatedly,
f
n
+5
=
f
n
+4
+
f
n
+3
=
(
f
n
+3
+
f
n
+2
) +
f
n
+3
=
2
f
n
+3
+
f
n
+2
=
(2
f
n
+2
+ 2
f
n
+1
) +
f
n
+2
=
3
f
n
+2
+ 2
f
n
+1
=
(3
f
n
+1
+ 3
f
n
) + 2
f
n
+1
=
5
f
n
+1
+ 3
f
n
≡
5
f
n
+1
+ 3
f
n
(mod 5)
≡
3
f
n
(mod 5)
Solution 2
We shall prove the result by strong induction on
n
.