Question

The mean of the cubes of the first $n$ natural numbers is :

• A. $\frac{n{(n+1)}^2}{2}$
• B. $\frac{n{(n+1)}^2}{4}$
• C. $n^2+n+1$
• D. $\frac{n(n+1)(n+2)}{8}$

Answer 1

The correct option is B

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

The explanation for the correct option.

The sum of cubes of first $n$ natural numbers is given by,

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

The mean of the group of observations is, dividing the sum of all observations by the number of observations.

Therefore, the mean of the cubes of the first $n$ natural numbers is computed as,

$$\begin{gathered} \therefore \mathrm{Mean}=\frac{{\displaystyle \frac{n^2{(n+1)}^2}{4}}}{n} \\ \Rightarrow =\frac{n^2{(n+1)}^2}{4n} \\ \Rightarrow =\frac{n{(n+1)}^2}{4} \end{gathered}$$

Hence, the option B is correct.