Question

The mean of the cubes of the first n natural numbers is?

Answer 1

Hint: We need to find the mean of the cubes of the first n natural numbers. We start to solve the given question by finding the sum of the cubes of the first n natural numbers. Then, we divide the sum of the cubes of the first n natural numbers by n to get the desired result.

Complete step-by-step answer:
We are asked to find out the mean of the cubes of the first n natural numbers. We will be solving the given question by finding the sum of the cubes of the first n natural numbers and then dividing the summation by n.
The concept of mean is widely used in mathematics and statistics. It is the average of two or more numbers.
The mean is also referred to as arithmetic mean and is given as follows,
$\Rightarrow mean={\displaystyle \frac{1}{n}}{\sum }_{i=1}^n{a}_i$
Here,
n is the number of values
$a_i$ is the data set values
Natural numbers are numbers that are greater than zero. They start from 1 and extend till infinity.
In mathematics, natural numbers are also referred to as counting numbers.
The first n natural numbers are 1, 2, 3, 4, ……, n
The sum of the cubes of first n natural numbers is given by
$\Rightarrow 1^3+2^3+3^3+...........+n^3$
From series and sequences,
We know that the sum of cubes of n natural numbers is given by ${\displaystyle \frac{n^2{(n+1)}^2}{4}}$
From the above, we get,
$\Rightarrow 1^3+2^3+3^3+...........+n^3={\displaystyle \frac{n^2{(n+1)}^2}{4}}$
Now, we need to find the mean of the cubes of the first n natural numbers.
We need to divide the above equation by n to find the mean.
Dividing the above equation by n on both sides, we get,
$\Rightarrow {\displaystyle \frac{1^3+2^3+3^3+...........+n^3}{n}}={\displaystyle \frac{n^2{(n+1)}^2}{4n}}$
Simplifying the above equation, we get,
$\therefore {\displaystyle \frac{1^3+2^3+3^3+...........+n^3}{n}}={\displaystyle \frac{n{(n+1)}^2}{4}}$

Note: Any mistake in writing the formula of the sum of cubes of n natural numbers will result in an incorrect solution. We must not forget to divide the sum of cubes of n natural numbers by n at the end of the problem.