Book of Proof - Amazon S3

278 Solutions
(2) Now assume the statement is true for some integer n = k ≥ 0, that is, assume
6 | (k 3 − k). This means k 3 − k = 6a for some integer a. We need to show that
6 | ((k + 1) 3 − (k + 1)). Observe that
(k + 1) 3 − (k + 1) = k 3 + 3k 2 + 3k + 1 − k − 1
= (k 3 − k) + 3k 2 + 3k
= 6a + 3k 2 + 3k
= 6a + 3k(k + 1).
Thus we have deduced (k +1) 3 −(k +1) = 6a +3k(k +1). Since one of k or (k +1)
must be even, it follows that k(k + 1) is even, so k(k + 1) = 2b for some integer
b. Consequently (k + 1) 3 − (k + 1) = 6a + 3k(k + 1) = 6a + 3(2b) = 6(a + b). Since
(k + 1) 3 − (k + 1) = 6(a + b) it follows that 6 | ((k + 1) 3 − (k + 1)).
Thus the result follows by mathematical induction.
15. If n ∈ N, then 1
1·2 + 1
2·3 + 1
3·4 + 1
4·5 + ··· + 1
n(n+1) = 1 − 1
n+1 .
Proof. The proof is by mathematical induction.
1
(1) When n = 1, the statement is
1(1+1) = 1 − 1
1+1 , which simplifies to 1 2 = 1 2 .
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
1
1·2 + 1
2·3 + 1
3·4 + 1
4·5 + ··· + 1
k(k+1) = 1 − 1
k+1
. Next we show that the statement for
n = k + 1 is true. Observe that
1
1 · 2 + 1
2 · 3 + 1
3 · 4 + 1
4 · 5 + ··· + 1
k(k + 1) + 1
(k + 1)((k + 1) + 1)
( 1
1 · 2 + 1
2 · 3 + 1
3 · 4 + 1
)
4 · 5 + ··· + 1
1
+
k(k + 1) (k + 1)(k + 2)
(
1 − 1 )
1
+
k + 1 (k + 1)(k + 2)
1 − 1
k + 1 + 1
(k + 1)(k + 2)
k + 2
1 −
(k + 1)(k + 2) + 1
(k + 1)(k + 2)
k + 1
1 −
(k + 1)(k + 2)
1 − 1
k + 2
1
1 −
(k + 1) + 1 .
This establishes 1
1·2 + 1
2·3 + 1
3·4 + 1
4·5 +···+ 1
(k+1)((k+1)+1 = 1− 1
that the statement is true for n = k + 1.
(k+1)+1
=
=
=
=
=
=
=
■
, which is to say
This completes the proof by mathematical induction.
■