Section 2.2 Induction
123
2. For all positive integers, let P(n) be the equation
2 + 4 + 6 + c + 2n = n(n + 1)
a. Write the equation for the base case P(1) and verify that it is true.
b. Write the inductive hypothesis P(k).
c. Write the equation for P(k + 1).
d. Prove that P(k + 1) is true.
In Exercises 3–26, use mathematical induction to prove that the statements are true for every positive integer n.
[Hint: In the algebra part of the proof, if the final expression you want has factors and you can pull those factors
out early, do that instead of multiplying everything out and getting some humongous expression.]
3. 1 + 5 + 9 + c + (4n − 3) = n(2n − 1)
4. 1 + 3 + 6 + c +
n(n + 1)
2
=
n(n + 1)(n + 2)
6
5. 4 + 10 + 16 + c + (6n − 2) = n(3n + 1)
6. 5 + 10 + 15 + c + 5n =
5n(n + 1)
2
7. 1
2
+ 2
2
+ c + n
2
=
n(n + 1)(2n + 1)
6
8. 1
3
+ 2
3
+ c + n
3
=
n
2
(n + 1)
2
4
9. 1
2
+ 3
2
+ c + (2n − 1)
2
=
n(2n − 1)(2n + 1)
3
10. 1
4
+ 2
4
+ c + n
4
=
n(n + 1)(2n + 1)(3n
2
+ 3n − 1)
30
11. 1 # 3 + 2 # 4 + 3 # 5 + c + n(n + 2) =
n(n + 1)(2n + 7)
6
12. 1 + a + a
2
+ c + a
n−1
=
a
n
− 1
a − 1
  for a ∙ 0, a ∙ 1
13.
1
1 # 2
+
1
2 # 3
+
1
3 # 4
+ c +
1
n(n + 1)
=
n
n + 1
14.
1
1 # 3
+
1
3 # 5
+
1
5 # 7
+ c +
1
(2n − 1)(2n + 1)
=
n
2n + 1
15. 1
2
− 2
2
+ 3
2
− 4
2
+ c + (−1)
n+1
n
2
=
(−1)
n+1
(n)(n + 1)
2
16. 2 + 6 + 18 + c + 2 # 3
n−1
= 3
n
− 1
17. 2
2
+ 4
2
+ c + (2n)
2
=
2n(n + 1)(2n + 1)
3
18. 1 # 2
1
+ 2 # 2
2
+ 3 # 2
3
+ c + n # 2
n
= (n − 1)2
n+1
+ 2
19. 1 # 2 + 2 # 3 + 3 # 4 + c + n(n + 1) =
n(n + 1)(n + 2)
3