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Example
Claim
:
Proof
by induction on
n
:
Base case (
n
= 0):
We need to show that
P
(0) is true i.e.,
Now
LHS = 0 (empty sum) and RHS = 0. Hence
P
(0) is true
Inductive step:
Let
k
∈
N
such that
P
(
k
)
is true i.e.,
We need to show that
Now
LHS =
=
RHS
Hence
P
(
k
+1) is true. The proof is now complete by induction.
2
)
1
(
,
1
+
=
Ν
∈
∀
∑
=
n
n
i
n
n
i
2
)
1
0
(
0
0
1
+
=
∑
=
i
i
2
)
1
(
1
+
=
∑
=
k
k
i
k
i
2
)
1
)
1
)((
1
(
1
1
+
+
+
=
∑
+
=
k
k
i
k
i
)
1
(
1
1
1
+
+
=
∑
∑
=
+
=
k
i
i
k
i
k
i
2
)
1
)
1
)((
1
(
)
1
(
2
)
1
(
+
+
+
=
+
+
+
=
k
k
k
k
k
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View Full DocumentAnother Example
Claim
:
∀
n
∈
N
,
3
(
n
3
−
n
)
Proof
by induction on
n
:
Base case (
n
= 0):
We need to show that
P
(0) is true i.e.,
3
(0
3
−
0)
Now (0
3
−
0) = 0
and since 0 = 0
×
3,
3
0.
Hence
P
(0) is true
Inductive step:
Let
k
∈
N
such that
P
(
k
)
is true i.e.,
3
(
k
3
−
k
)
[ IH ]
We need to show that
3
((
k
+1)
3
−
(
k
+1))
Now
((
k
+1)
3
−
(
k
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