Question

Let S(n) = 1 + 2 + · · · + n be the sum of the first n natural numbers and let$\mathit{C}\left(n\right)\mathbf{=}{\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{2}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{n}}^{\mathbf{3}}$be the sum of the first n cubes. Prove the following equalities by induction on n, to arrive at the curious conclusion that ${\mathit{C}}_{\mathbf{n}}\mathbf{=}{\left(S_n\right)}^{\mathbf{2}}$ for every n.

a. $\mathit{S}\left(n\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{n}\left(n+1\right)$

b.$\mathit{C}\left(n\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left(n^4+2n^3+n^2\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}{\mathit{n}}^{\mathbf{2}}{\left(n+1\right)}^{\mathbf{2}}$

Short answer

a. The given equality of $S\left(n\right)=\frac{1}{2}n\left(n+1\right)$ can be proved by induction.

b. The given equality of$C\left(n\right)=\frac{1}{4}\left(n^4+2n^3+n^2\right)=\frac{1}{4}{n}^2{\left(n+1\right)}^2$

Step-by-step solution

Explain given information:

Given is the sum of first natural numbers as $S\left(n\right)=1+2+\dots +n$and the sum of first ncubes be$C\left(n\right)=1^3+2^3+\dots +n^3$ . The equalities has to be proved by the induction method.

(a) Prove equality by induction.

Consider that the sum of first natural numbers $S\left(n\right)=1+2+\dots +n$be,

$S\left(n\right)=\frac{1}{2}n\left(n+1\right)$

By induction method, rewrite $S\left(n\right)=1+2+\dots +n$as, $\sum _{i=1}^{n}i=\frac{1}{2}n\left(n+1\right)$

For $n=1,$

$$\begin{gathered} 1=\frac{1}{2}1\left(1+1\right) \\ 1=\frac{2}{2}=1 \end{gathered}$$

It is true.

For $n=2,$

$$\begin{gathered} 1+2=\frac{1}{2}2\left(2+1\right) \\ 2=\frac{6}{2}=2 \end{gathered}$$

It is true.

Thus, it is true for $n$.

Prove it for $n+1$,$S_{n+1}=\sum _{i=1}^{n+1}i=\frac{1}{2}\left(n+1\right)\left(n+1\right)$

$\begin{array}{rcl}{S}_{n+1}& =& 1+2+\dots +n+n+1\\ & =& S_n+n+1\\ & =& \frac{n\left(n+1\right)}{2}+n+1\\ & =& \frac{\left(n+1\right)\left(n+2\right)}{2}\end{array}$

$S_{n+1}=\sum _{i=1}^{n+1}i=\frac{1}{2}\left(n+1\right)\left(n+2\right)$

Therefore, it has been proved that the given equality for sum of natural numbers is also true for $n+1$.Hence the given equality is correct.

(b) Prove equality by induction.

Consider the sum of the cube $C\left(n\right)=1^3+2^3+\dots +n^3$is given by,

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

The given equality can be written as,

$\sum _{i=1}^{k=n}{i}^3=1^3+2^3+....+n^3=\frac{1}{4}{n}^2{\left(n+1\right)}^2$

For$n=1$ ,

$\begin{array}{rcl}{1}^3& =& \frac{1}{4}{1}^2{\left(1+1\right)}^2\\ 1& =& 1\end{array}$

It is true.

It is true for$n=n$ also.

For, $n=n+1$,

$\begin{array}{rcl}{C}_{n+1}& =& 1^3+2^3+...+n^3+{\left(n+1\right)}^3\\ & =& \frac{1}{4}{n}^2{\left(n+1\right)}^2+{\left(n+1\right)}^3\\ & =& {\left(n+1\right)}^2\left(\frac{n^2+4n+4}{4}\right)\\ & =& \frac{{\left(n+1\right)}^2{\left(n+2\right)}^2}{4}\end{array}$

Therefore, the given equality is true for n+1, hence it is correct.

From the above explanation,

$\begin{array}{rcl}{C}_n& =& \frac{1}{4}{n}^2{\left(n+1\right)}^2={\left[\frac{n\left(n+1\right)}{2}\right]}^2\\ C_n& =& {\left(S_n\right)}^2\end{array}$

Therefore, it is concluded that $C_n={\left(S_n\right)}^2$