Rosen 3.3
p253.
1. Sum of the first n even natural numbers = 2 + 4 + 6 + ... + 2n = 2[1 + 2 + 3 + ... + n]
=
for n $ 0.
3. Prove P(n) :
Basis: for n = 0, it is clear that
; so P(0) is true.
Induction: Show P(k)YP(k+1) for k $ 0: Given
, for k $ 0,
consider
which
.
Now because P(0) is true and P(k)YP(k+1) for k $ 0, P(n) is true for all n $ 0.
, for n $ 1.
5. The expression
Basis: for n = 1, it is clear that
; sp P(1) is true.
Induction: Show P(k)YP(k+1) for k $ 1: Given
consider
, for k $ 1,
which
.
Now because P(1) is true and P(k)YP(k+1) for k $ 1, P(n) is true for all n $ 1.
13. Show that 2n > n2 whenever n is greater than 4.
Basis: 25 = 32 > 25 = 52, so P(5) is true.
Induction: Show P(k)YP(k+1) for k $ 4: Given 2k > k2 for k > 4,
consider 2(k+1) = 2k@2 = 2k + 2k which is > k2 + k2 which is > k2 + 4k for k > 4
and k2 + 4k = k2 + (2k + 2k) which is > (k2 + 2k) + 8 for k > 4
and (k2 + 2k) + 8 > (k2 + 2k) + 1 which = (k+1)2. Thus, 2(k+1) > (k+1)2 for k > 4.
Now because P(5) is true and P(k)YP(k+1) for k > 4, P(n) is true for all n > 4.
21. P(n) : 5|(n5 – n) for n $ 0.
Basis: 05 – 0 = 0 = 5@0, so it is clear that P(0) is true.
Induction: Show P(k)YP(k+1) for k $ 0 : Given 5 | (k5 – k) for k $ 0, we have (k5 – k) = 5q for k $ 0.
Consider (k+1)5 – (k+1) = k5 + 5k4 + 10k3 + 10k2 + 5k + 1 – k – 1
which = (k5 – k) + 5k4 + 10k3 + 10k2 + 5k = 5q + 5k4 + 10k3 + 10k2 + 5k = 5[q + k4 + 2k3 + 2k2 + k].
Now because P(0) is true and P(k)YP(k+1) for k $ 0, P(n) is true for all n $ 0.