Question

The arithmetic mean of the square of the first n natural number is
A. $\dfrac{n(n+1)(2n+1)}{6}$
B. $\dfrac{n(n+1)(2n+1)}{2}$
C. $\dfrac{(n+1)(2n+1)}{6}$
D. $\dfrac{(n+1)(2n+1)}{2}$

Answer 1

Hint: In general arithmetic mean or average of a set of values in the ratio of the sum of these values to the number of elements in the set. Otherway, we add together the given values in a data set and then divide that total by the number of given values.

Complete step-by-step solution -
If $\mathop{x}_{1},\mathop{x}_{2},\mathop{x}_{3}.........,\mathop{x}_{n}$are n observations, then their arithmetic mean is given by-
$\Rightarrow \overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+........+{{x}_{4}}}{n}$
$\Rightarrow \overline{x}=\sum\limits_{i=1}^{n}{\dfrac{{{x}_{i}}}{n}}$
Let, m be the arithmetic mean of two quantities x and y.
Then $x,m,y$are in the arithmetic progression
Now, $m-x=y-m=$common difference
Therefore, the arithmetic mean between any two given quantities is half – their sum.
If more than three terms are in arithmetic progression, then the terms between the two extremes are called the arithmetic means between the extreme terms.
Natural numbers are the positive integers (whole number)
Example -$1,2,3....,n$ etc.
The first n natural numbers are $1,2,3....,n$
Sum of squares of n natural numbers is $\mathop{1}^{2}+\mathop{2}^{2}+\mathop{3}^{2}+.......\mathop{n}^{2}$
Let us assume the required Sum = S
Therefore, $S=\mathop{1}^{2}+\mathop{2}^{2}+\mathop{3}^{2}+\mathop{4}^{2}+\mathop{5}^{2}+.........+\mathop{n}^{2}$
Now, we will use the below identity to find the value of squares of n natural number (S)
$\mathop{n}^{3}-\mathop{(n-1)}^{3}=3\mathop{n}^{2}-3n+1$
Substituting, $n=1,2,3,4,5........,n$ in the above, identity
We get,
$\mathop{\Rightarrow 1}^{3}-\mathop{0}^{3}={{3.1}^{2}}-3.1+1$
$\mathop{\Rightarrow 2}^{3}-\mathop{1}^{3}={{3.2}^{2}}-3.2+1$
$\mathop{\Rightarrow 3}^{3}-\mathop{2}^{3}={{3.3}^{2}}-3.3+1$
$\mathop{\Rightarrow 4}^{3}-\mathop{3}^{3}={{3.4}^{2}}-3.4+1$
$....................................$
$\mathop{\Rightarrow n}^{3}-\mathop{(n-1)}^{3}=3.{{n}^{2}}-3.n+1$
Adding we get,
$\mathop{n}^{3}-\mathop{0}^{3}=3(\mathop{1}^{2}+\mathop{2}^{2}+\mathop{3}^{2}+\mathop{4}^{2}+.........+\mathop{n}^{2})-3(1+2+3+4+.............+n)+(1+2+1+1+1+.......\text{n times})$
$\Rightarrow \mathop{n}^{3}=3s-3.\dfrac{n(n+1)}{2}+n$ [Sum of the first n natural number $=1+2+3+.......+n=\dfrac{n(n+1)}{2}$]
$\Rightarrow 3s=\mathop{n}^{3}+\dfrac{3}{2}n(n+1)-n=(\mathop{n}^{2}-1)+\dfrac{3}{2}n(n+1)$
$\Rightarrow 3s=n(n+1)\left( n-1+\dfrac{3}{2} \right)$
$\Rightarrow 3x=n(n+1)\left( \dfrac{2n-2+3}{2} \right)$
$\Rightarrow 3s=\dfrac{n(n+1)(2n+1)}{2}$
Therefore,$S=\dfrac{n(n+1)(2n+1)}{6}$
Such that, $\mathop{1}^{2}+\mathop{2}^{2}+\mathop{3}^{2}+\mathop{4}^{2}+\mathop{5}^{2}+.........+\mathop{n}^{2}=\dfrac{n(n+1)(2n+1)}{6}$
Thus, the sum of the square first n natural numbers $'S'=\dfrac{n(n+1)(2n+1)}{6}$
We know that, the sum of the squares of first n natural numbers is given by-
$\dfrac{n(n+1)(2n+1)}{6}$
Arithmetic mean is the average of values of the data set.
Arithmetic mean (A.M) $=\dfrac{\text{sum of total observation}}{\text{Total no}\text{. of observation}}$
$\text{A}\text{.M}\text{.}=\dfrac{\text{sum of the squares of first n natural number}}{\text{Total no}\text{. of observation}}$
[As, the sum of the squares of first n natural number is $(S)=\dfrac{n(n+1)(2n+1)}{6}$]
n= Total no. of observation
Mean $=\dfrac{\mathop{1}^{2}+\mathop{2}^{2}+\mathop{3}^{2}+.........+\mathop{n}^{2}}{n}$
$\Rightarrow A.M=\dfrac{\sum{\mathop{n}^{2}}}{n}$
$\Rightarrow A.M=\dfrac{n(n+1)(2n+1)}{6\times n}$
$\Rightarrow A.M=\dfrac{(n+1)(2n+1)}{6}$
Hence option c is correct.

Note: We clear the arithmetic mean and the sum of squares of first n natural numbers to avoid any confusion regarding this.
First, we have to prove the sum of squares of the first n natural number, then with this help we calculate the arithmetic mean.