Question

Let P(n)be the statement that ${\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{2}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{n}}^{\mathbf{3}}\mathbf{=}\mathbf{\left(}\mathit{n}\mathbf{\right(}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{/}\mathbf{2}{\mathbf{)}}^{\mathbf{2}}$ for the positive integer .

a) What is the statement P(1)?

b) Show that P(1) is true, completing the basis step of

the proof.

c) What is the inductive hypothesis?

d) What do you need to prove in the inductive step?

e) Complete the inductive step, identifying where you

use the inductive hypothesis.

f) Explain why these steps show that this formula is true wheneveris a positive integer.

Short answer

(a) The statement P(1) is, $1^3={\left(\frac{1(1+1)}{2}\right)}^2$.

(b) It is shown that the statement P(1) is true.

(c) The assumption that P(k)is true is called the inductive hypothesis.

(d) In the inductive step, we need to prove that, if P(k) is true, then P(K+1) is also true.

(e) P(k+1)is true under the assumption that P(k)is true. This

completes the inductive step.

(f) Since we have shown that the formula is true for P(1) and P(K) is true for any arbitrary integer K and also for K + 1 . This implies it is true for any positive integer.

Step-by-step solution

Principle of Mathematical Induction

To prove that P(n)is true for all positive integers n, where P(n)is a propositional function, we complete two steps:

Basis Step:We verify that P(1)is true.

Inductive Step:We show that the conditional statement $\mathit{P}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{\to }\mathit{P}\mathbf{(}\mathit{k}\mathbf{+}\mathbf{1}\mathbf{)}$is true for all positive integers k.

(a) Step 2: Finding statement P(n) 

Given statement is

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

For finding statement P(1) substituting 1 for n in the statement

Therefore, the statement P(1) is

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

(b) Step 3: Showing P(1)  is true

Given statement is

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

For finding statement P(1) substituting 1 for n in the statement

$\begin{array}{r}{1}^3={\left(\frac{1(1+1)}{2}\right)}^2\\ 1={\left(\frac{2}{2}\right)}^2\\ 1=1^2\\ 1=1\end{array}$

Therefore, the statement P(1) is true this is also known as the basis step of the proof.

(c) Step 4: Inductive hypothesis

The assumption that P(K)is true is called the inductive hypothesis.

Given statement is

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

Therefore, the inductive hypothesis is

$1^3+2^3+\dots +k^3={\left(\frac{k(k+1)}{2}\right)}^2$

(d) Step 5: Inductive step

In the inductive step, we need to prove that, if P(K) is true, then P ( K + 1 ) is also true.

That is,

$P\left(k\right)\to P\left(K+1\right)$is true for all positive integers k.

(e) Step 6: Completing the inductive step

In the inductive hypothesis, we assume that P(K) is true for any arbitrary positive integer

That is

$1^3+2^3+\dots +k^3={\left(\frac{k(k+1)}{2}\right)}^2$…(i)

Now we must have to show that P(K+1) is also true

Therefore replacing K with K + 1 in the statement

$\begin{array}{r}{1}^3+2^3+\dots +(k+1{)}^3={\left(\frac{(k+1)\left(\right(k+1)+1)}{2}\right)}^2\\ 1^3+2^3+\dots +(k+1{)}^3={\left(\frac{(k+1)(k+2)}{2}\right)}^2\end{array}$

Now, Adding${\left(K+1\right)}^3$ in both sides of the equation (i) or inductive hypothesis.

$\begin{array}{r}{1}^3+2^3+\dots +k^3+(k+1{)}^3={\left(\frac{k(k+1)}{2}\right)}^2+(k+1{)}^3\\ =\frac{\left(k\right(k+1){)}^2}{4}+\frac{4(k+1{)}^3}{4}\\ =\frac{(k+1{)}^2\left(\left(k^2\right)+4(k+1)\right)}{4}\mid \\ =\frac{(k+1{)}^2\left(k^2+4k+1\right)}{4}\\ =\frac{(k+1{)}^2(k+2{)}^2}{4}\\ {\left.=\frac{(k+1)(k+2)}{2}\right)}^2\end{array}$

From the above, we can see that P (K+1) is also true

Hence, P(K+1) is true under the assumption that P(K)is true. This

completes the inductive step.

(f) Explanation that this formula is true whenever  is a positive integer.

Since we have shown that the formula is true for P(1) and P(K) is true for any arbitrary integer K and also for K + 1 . This implies it is true for any positive integer.

Hence this formula is true whenever n is a positive integer.