Question

${(1^r+2^r+3^r+\dots .+n^r)}^n$is

• A. $>n^n$
• B. $>n^n\left(n\right)!$
• C. $>n^n{(n!)}^{r}forallr$
• D. None of these

Answer 1

The correct option is C

$>n^n{(n!)}^{r}forallr$

Explanation for the correct option:

Solving the given equation:

Given, ${(1^r+2^r+3^r+\dots .+n^r)}^n$

Here, $1,2,3,...$so on are non equal positive integers

So, $\frac{(1^r+2^r+3^r+\dots .+n^r)}{n}>{(1^r\times 2^r\times 3^r\times \dots .\times n^r)}^{\frac{1}{n}}$ $\left[\mathbf{\because }\mathit{A}\mathit{M}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{n}\mathbf{}\mathit{q}\mathit{u}\mathit{a}\mathit{n}\mathit{t}\mathit{i}\mathit{t}\mathit{i}\mathit{e}\mathit{s}\mathbf{}\mathbf{>}\mathbf{}\mathit{G}\mathit{M}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{n}\mathbf{}\mathit{q}\mathit{u}\mathit{a}\mathit{n}\mathit{t}\mathit{i}\mathit{t}\mathit{i}\mathit{e}\mathit{s}\right]$

or, ${(1^r+2^r+3^r+\dots .+n^r)}^n>n^n{(1\times 2\times 3.\dots ..n)}^r$

or, ${\mathbf{(}{\mathbf{1}}^{\mathbf{r}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{2}}^{\mathbf{r}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{3}}^{\mathbf{r}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{\dots }\mathbf{.}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{n}}^{\mathbf{r}}\mathbf{)}}^{\mathbf{n}}\mathbf{}$$\mathbf{>}\mathbf{}{\mathit{n}}^{\mathbf{n}}\mathbf{}\mathbf{(}\mathit{n}\mathbf{!}{\mathbf{)}}^{\mathbf{r}}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{r}$

Hence, Option ‘C’ is Correct.