Verify that for all n≥ 1 the sum of the squares of the first 2n positive integers is given by
1²+2²+3²+...+ (2n)² = n (2n+1) (4n+1)/3
Answers
Answer:Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despi´c
In Exercises 1-15 use mathematical induction to establish the formula for n ≥ 1.
1. 1
2 + 22 + 32 + · · · + n
2 =
n(n + 1)(2n + 1)
6
Proof:
For n = 1, the statement reduces to 12 =
1 · 2 · 3
6
and is obviously true.
Assuming the statement is true for n = k:
1
2 + 22 + 32 + · · · + k
2 =
k(k + 1)(2k + 1)
6
, (1)
we will prove that the statement must be true for n = k + 1:
1
2 + 22 + 32 + · · · + (k + 1)2 =
(k + 1)(k + 2)(2k + 3)
6
. (2)
The left-hand side of (2) can be written as
1
2 + 22 + 32 + · · · + k
2 + (k + 1)2
.
In view of (1), this simplifies to:
1
2 + 22 + 32 + · · · + k
2
+ (k + 1)2 =
k(k + 1)(2k + 1)
6
+ (k + 1)2
=
k(k + 1)(2k + 1) + 6(k + 1)2
6
=
(k + 1)[k(2k + 1) + 6(k + 1)]
6
=
(k + 1)(2k
2 + 7k + 6)
6
=
(k + 1)(k + 2)(2k + 3)
6
.
Thus the left-hand side of (2) is equal to the right-hand side of (2). This
proves the inductive step. Therefore, by the principle of mathematical
induction, the given statement is true for every positive integer n.
2. 3 + 32 + 33 + · · · + 3n =
3
n+1 − 3
2
Proof:
For n = 1, the statement reduces to 3 = 3
2 − 3
2
and is obviously true.
Assuming the statement is true for n = k:
3 + 32 + 33 + · · · + 3k =
3
k+1 − 3
2
, (3)
we will prove that the statement must be true for n = k + 1:
3 + 32 + 33 + · · · + 3k+1 =
3
k+2 − 3
2
. (4
Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despi´c
The left-hand side of (4) can be written as
3 + 32 + 33 + · · · + 3k + 3k+1
.
In view of (3), this simplifies to:
3 + 32 + 33 + · · · + 3k
+ 3k+1 =
3
k+1 − 3
2
+ 3k+1
=
3
k+1 − 3 + 2 · 3
k+1
2
=
3 · 3
k+1 − 3
2
=
3
k+2 − 3
2
.
Thus the left-hand side of (4) is equal to the right-hand side of (4). This
proves the inductive step. Therefore, by the principle of mathematical
induction, the given statement is true for every positive integer n.
3. 1
3 + 23 + 33 + · · · + n
3 =
n
2
(n + 1)2
4
Proof:
For n = 1, the statement reduces to 13 =
1
2
· 2
2
4
and is obviously true.
Assuming the statement is true for n = k:
1
3 + 23 + 33 + · · · + k
3 =
k
2
(k + 1)2
4
, (5)
we will prove that the statement must be true for n = k + 1:
1
3 + 23 + 33 + · · · + (k + 1)3 =
(k + 1)2
(k + 2)2
4
. (6)
The left-hand side of (6) can be written as
1
3 + 23 + 33 + · · · + k
3 + (k + 1)3
.
In view of (5), this simplifies to:
1
3 + 23 + 33 + · · · + k
3
+ (k + 1)3 =
k
2
(k + 1)2
4
+ (k + 1)3
=
k
2
(k + 1)2 + 4(k + 1)3
4
=
(k + 1)2
[k
2 + 4(k + 1)]
4
=
(k + 1)2
(k + 2)2
4
.
Thus the left-hand side of (6) is equal to the right-hand side of (6). This
proves the inductive step. Therefore, by the principle of mathematic
Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despi´c
induction, the given statement is true for every positive integer n.
Step-by-step explanation: