Graphics Reference
In-Depth Information
Answers and comments
1
Let
P
(
n
) mean '1
2
3
...
n
n
(
n
1)(2
n
1)/6'.
You checked that
P
(1),
P
(2) and
P
(3) were all true.
The algebra you did showed that
P
(
n
)
1).
So (by induction)
P
(
n
) is truefor all natural numbrs
n
.
P
(
n
2
Let
P
(
n
) mean '6
5
n
4 is divisibleby 5'.
Because 5
5,
P
(1) is true. Because 5
30,
P
(2) is true. Because 5
205,
P
(3) is true.
The difference between the expression for
n
1 and the expression for
n
is 5 ยท 6
5, which is divisibleby 5, so
P
(
n
)
P
(
n
1). And (by
induction)
P
(
n
) is truefor all natural numbrs
n
.
3
Takeeach of thesix propositions in turn as
P
(
n
). Verify the truth of
P
(1) in each case. The algebra behind the proof of
P
(
n
)
P
(
n
1) is as
follows:
(i)
n
(
n
1)
(
n
1)
(
n
1)(
n
2),
(ii)
n
[2
a
(
n
1)
d
]
(
a
nd
)
(
n
1)(2
a
nd
),
n
(
n
(
n
(
n
1)(
n
(iii) [
1)]
1)
[
2)]
,
(iv)
n
(
n
1)(
n
2)
(
n
1)(
n
2)
(
n
1)(
n
2)(
n
3),
(v) (2
1)
2
2
1,
(vi)
x
1
x
1
1
x
.
x
x
1
nx
x
1
nx
(
n
1)
x
(
x
1)
x
1
(vii)
1
(
n
1)
x
x
(
x
1)
x
1
(
x
1)
(
n
1)
x
x
1
1
.
x
(
x
1)
4
If
P
(
n
) is theproposition weareasked to provefor 0
r
n
,
n
0
n
n
1
is easily checked, and incorporates the truth of
P
(1). If weassume
P
(
n
)
and apply it to the Pascal triangle property, we get
P
(
n
1).
Symmetry about a vertical axis.
5
For the inductive step, one must show that the coeMcient of
x
in
n
n
r
(1
x
)
(1
x
)
(1
x
)is
.
r
1
