Solutions for exercises in chapter 1 E1.1 Prove by induction that for ...

For n = 1, the left side is 1 · 2 = 2, while the right side is 1·2·3
3
that the equation holds for n. Then
= 6
3
= 2 also. Now assume
n(n + 1)(n + 2)
1 · 2 + 2 · 3 + · · · + n · (n + 1) + (n + 1) · (n + 2) = + (n + 1)(n + 2)
3
= n(n + 1)(n + 2) + 3(n + 1)(n + 2)
3
= (n + 1)(n + 2)(n + 3)
,
3
which gives the equation with n replaced by n + 1. So the equation holds for all positive
integers n.
E1.4 Recall that for m a positive integer, m! = m · (m − 1) · . . . · 1, with 0! = 1; and for
0 ≤ i ≤ m, � �
m
=
i
Prove that if 1 ≤ i ≤ m, then
(Induction is not needed.)
� � � �
m m
+ =
i i − 1
m!
i!(m − i)! .
� � � �
m m
+ =
i i − 1
= m!(m − i + 1)
m!
i!(m − i)! +
� m + 1
i!(m − i + 1)! +
= m!(m − i + 1 + i)
i!(m − i + 1)!
= (m + 1)!
i!(m + 1 − i)!
� �
m + 1
= .
i
i
�
.
m!
(i − 1)!(m − i + 1)!
m!i
i!(m − i + 1)!
E1.5 Prove by induction the binomial theorem, that for any n ∈ ω and any real numbers
a, b,
(a + b) n n�
=
i=0
For n = 0 the left side is 1, while the right side is
0�
i=0
� �
0
a
i
i b 0−i =
2
� �
n
a
i
i b n−i ,
� �
0
a
0
0 b 0 = 1.