3. We prove the result by strong induction on
n
.
Base Cases
When
n
= 1, we have
1
2
(2
n
+ (

2)
n
) =
1
2
(2

2) = 0 =
y
1
.
When
n
= 2, we have
1
2
(2
n
+ (

2)
n
) =
1
2
(4 + 4) = 4 =
y
2
.
Therefore, the result holds when
n
= 1 and
n
= 2.
Induction Hypothesis
Suppose that the result holds for
n
= 1
,
2
, . . . , k
, for some
k
∈
P
,
k
≥
2.
That is, suppose that
y
n
=
1
2
(2
n
+ (

2)
n
) for
n
= 1
,
2
, . . . , k
.
Induction Conclusion
Consider
n
=
k
+ 1. Then
y
k
+1
= 4
y
k

1
= 4
(
1
2
(2
k

1
+ (

2)
k

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

1
+ (

2)
k

1
)
=
1
2
(2
2
2
k

1
+ (

2)
2
(

2)
k

1
)
=
1
2
(2
k
+1
+ (

2)
k
+1
)
so the result holds for
n
=
k
+ 1.
Therefore, the result holds for all
n
∈
P
by POSI.
4. We prove the result by strong induction on
n
.
Base Cases
When
n
= 1, we have 2(

1)
n

(

2)
n
+ 3
n
= 2(

1)

(

2) + 3 = 3 =
t
1
.
When
n
= 2, we have 2(

1)
n

(

2)
n
+ 3
n
= 2(1)

4 + 9 = 7 =
t
2
.
When
n
= 3, we have 2(

1)
n

(

2)
n
+ 3
n
= 2(

1)

(

8) + 27 = 33 =
t
3
.
Induction Hypothesis
Suppose the result is true for
n
= 1
,
2
, . . . , k
, for some
k
∈
P
,
k
≥
3. That is, suppose
t
n
= 2(

1)
n

(

2)
n
+ 3
n
for
n
= 1
,
2
, . . . , k
.
Induction Conclusion
Consider
n
=
k
+ 1. Then
t
k
+1
= 7
t
k

1
+ 6
t
k

2
= 7(2(

1)
k

1

(

2)
k

1
+ 3
k

1
) + 6(2(

1)
k

2

(

2)
k

2
+ 3
k

2
)
= 14(

1)
k

1

7(

2)
k

1
+ 7(3
k

1
) + 12(

1)
k

2

6(

2)
k

2
+ 6(3
k

2
)
= (

1)
k

2
[14(

1) + 12]

(

2)
k

2
[7(

2) + 6] + 3
k

2
[7(3) + 6]
= (

1)
k

2
[

2]

(
