Answer to Question #88442 in Discrete Mathematics for MIQUELLE CORDOVA

Question

4. Use mathematical induction to prove that 1^3 + 2^3 + ... + n^3 =
=(n(n+1)/2)^2
for all integers n ≥ 1

Answer 1

According to the method of mathematical induction, one has to prove that statement

1) is correct for the initial value (in this task it is n=1)

2) assuming that the statement is valid for arbitrary n, prove its validity for (n+1).

Executing these steps, we obtain:

1) n = 1:

$1^3 = 1, \quad \left( \frac{1 \cdot 2}{2}\right)^2 = 1, \quad 1=1$

statement is proved.

2) assuming that

$1^3 + 2^3 + ... + n^3 = \left( \frac{n(n+1)}{2}\right)^2$

is correct, let us check it for (n+1) case:

$1^3 + 2^3 + ... + n^3 + (n+1)^3 \stackrel{?}{=} \left( \frac{(n+1)(n+2)}{2}\right)^2$

Simplifying the left-hand side of the expression, one can derive:

$(1^3 + 2^3 + ... + n^3) + (n+1)^3 = \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 =$

$=(n+1)^2 \left(\frac{n^2}{4} + (n+1) \right) = (n+1)^2 \frac{n^2 +4n+4}{4} = \left(\frac{(n+1)(n+2)}{2}\right)^2,$

which coincides with the right-hand side of the assumption.

By the method of mathematical induction the statement is true for all natural values of n.