Question

Prove by mathematical induction the formula (1^3+2^3+3^3+4^3+....n^3)=(n^2(n+1)^2)/4

Answer 1

Answer on Question #85906 – Math – Discrete Mathematics

Question

Prove by mathematical induction the formula

$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = \frac{n^2(n + 1)^2}{4}$$

Solution

For any integer $n \geq 1$, let $P_n$ be the statement that

$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = \frac{n^2(n + 1)^2}{4}$$

Base case. The statement $P_1$ says that

$$1^3 = \frac{1^2(1 + 1)^2}{4},$$

which is true.

Inductive case. Fix $k \geq 1$, and suppose that $P_k$ holds, that is,

$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 = \frac{k^2(k + 1)^2}{4}$$

It remains to show that $P_{k+1}$ holds, that is,

$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + (k + 1)^3 = \frac{(k + 1)^2((k + 1) + 1)^2}{4}$$ $$\begin{array}{l} 1^3 + 2^3 + 3^3 + 4^3 + \cdots + (k + 1)^3 = \\ = 1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 + (k + 1)^3 = \\ = \frac{k^2(k + 1)^2}{4} + (k + 1)^3 = \\ = (k + 1)^2\left(\frac{k^2}{4} + k + 1\right) = \\ = \frac{(k + 1)^2}{4} (k^2 + 4k + 4) = \\ = \frac{(k + 1)^2}{4} (k + 2)^2 = \\ = \frac{(k + 1)^2((k + 1) + 1)^2}{4} \end{array}$$

Therefore, $P_{k+1}$ holds.

Therefore, by the principle of mathematical induction, for all $n \geq 1$, $P_n$ holds

$$1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = \frac{n^2(n + 1)^2}{4}, n \in \mathbb{N}.$$

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