Question

Use mathematical induction to show that

${\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{3}}^{\mathbf{3}}\mathbf{+}{\mathbf{5}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}{\mathbf{)}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\left(2n^2+4n+1\right)$whenever nis a positive integer.

Short answer

It is shown that $1^3+3^3+5^3+\dots +(2\mathrm{n}+1{)}^3=(\mathrm{n}+1{)}^2(2{\mathrm{n}}^2+4\mathrm{n}+1)$ whenevernis a positive integer.

Step-by-step solution

Principle of Mathematical Induction

Consider the propositional function P (n) . Consider two actions to prove that P (n) evaluates to accurate for all set of positive integers .

Consider the first basic step is to confirm thatP (1) true.

Consider the inductive step is to demonstrate that for any positive integer k the conditional statement $\mathit{P}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{\to }\mathit{P}\mathbf{(}\mathit{k}\mathbf{+}\mathbf{1}\mathbf{)}$is true.

Prove the basis step

Given statement is

$1^3+3^3+5^3+\dots +(2n+1{)}^3=(n+1{)}^2\left(2n^2+4n+1\right)$

In the basis step, we need to prove that is true.

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

Therefore, the statement P (1) is

$$\begin{gathered} 1^3+3^3+5^3+\dots +(2\mathrm{n}+1{)}^3=(\mathrm{n}+1{)}^2\left(2{\mathrm{n}}^2+4\mathrm{n}+1\right) \\ \begin{array}{r}{1}^3+\left(2\right(1)+1{)}^3=(1+1{)}^2\left(2(1{)}^2+4(1)+1\right)\\ 1+27=4\left(7\right)\\ 28=28\end{array} \end{gathered}$$

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

Prove the 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,

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

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

That is

$1^3+3^3+5^3+\dots +(2k+1{)}^3=(k+1{)}^2\left(2k^2+4k+1\right)$

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

Therefore replacing k with k + 1 in the statement

$$" width="9" height="19" role="math">13+33+53+…+(2k+1)3=(k+1)22k2+4k+1

Now, Adding $\left(2\right(k+1)+1{)}^3$in both sides of the equation (i) or inductive hypothesis.

$\left(2\right(k+1)+1{)}^3\begin{array}{r}{1}^3+3^3+5^3+\dots +(2k+1{)}^3+(2(k+1)+1{)}^3=(k+1{)}^2\left(2k^2+4k+1\right)+(2(k+1)+1{)}^3\\ =2k^4+16k^3+47k^2+60k+28\\ =(k+2{)}^2\left(2k^2+8k+7\right)\\ =\left(\right(k+1)+1{)}^2\left(2(k+1{)}^2+4(k+1)+1\right)\end{array}$

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.

Hence, it is proved that $1^3+3^3+5^3+\dots +(2\mathrm{n}+1{)}^3=(\mathrm{n}+1{)}^2\left(2{\mathrm{n}}^2+4\mathrm{n}+1\right)$whenevernis a positive integer.