Prove by mathematical induction that 1^3 + 2^3 + 3^3 + · · · + n^3 = (n(n + 1)/2)^2 for all n ≥ 1.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Exercise 1.2. Prove by mathematical induction that
1^3 + 2^3 + 3^3 + · · · + n^3 = (n(n + 1)/2)^2
for all n ≥ 1.
Proof. We first abbreviate
13 + 23 + 33 + · · · + n3 = Xn
m=1
m3.
So, we need to prove for all n ≥ 1,
Xn
m=1
m3 =
n(n + 1)
2
2
.
We will proceed by mathematical induction.
(i) We need to check that the above equation holds when n = 1.
LHS: P1
m=1 m3 = 13 = 1.
RHS: 1·(1+1)
2
2
= � 1·2
2
2 = 12 = 1.
Thus, the equality holds when n = 1.
2
(ii) Let k ∈ N, n > 1. Assume by induction hypothesis that the equality holds when n = k. That is, assuming
X
k
m=1
m3 =
k(k + 1)
2
2
is our induction hypothesis. We next attempt to prove the desired equality for k + 1:
k
X
+1
m=1
m3 = 13 + 23 + 33 + · · · + k3 + (k + 1)3
= X
k
m=1
m3 + (k + 1)3
=
k(k + 1)
2
2
+ (k + 1)3
= (k + 1)2 ·

k2
22 + (k + 1)
= (k + 1)2 ·
k2 + 4(k + 1)
4
= (k + 1)2(k2 + 4k + 4)
4
=

(k + 1)(k + 2)
2
2
=

(k + 1)((k + 1) + 1)
2
2
,
as desired.
According to the principle of induction, since (i) and (ii) hold, we have the given equality is valid for all n ≥ 1.
This concludes our proof.
SOLVE this exercise in this way:
establish this formulas below by mathematical induction:
1^3 + 2^3 + 3^3 + · · · + n^3 = [n(n + 1)/2]^2

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