Question

The mean of the squares of the first $ n $ natural numbers is?
A. $ {{n}^{2}}+1 $
B. $ \dfrac{{{n}^{4}}+1}{n} $
C. $ \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} $
D. $ \dfrac{\left( n+1 \right)\left( n+2 \right)}{m} $

Answer 1

Hint: In this problem we will assume the $ n $ natural numbers and we will calculate the squares of the $ n $ natural numbers and then we will calculate the value of sum of squares of the $ n $ natural numbers. We know that the average is the ratio of sum of the variables to the total number of variables. So, we will find the ratio of the sum of squares of the $ n $ natural numbers to the $ n $ natural numbers.

Complete step-by-step answer:
Let the $ n $ natural numbers are $ 1,2,3,4,...,n $
Now the squares of the $ n $ natural numbers are $ {{1}^{2}},{{2}^{2}},{{3}^{2}},{{4}^{2}},...,{{n}^{2}} $
The sum of the squares of the $ n $ natural numbers is $ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{n}^{2}} $ .
Let the sum of the square of the $ n $ natural numbers is
 $ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{n}^{2}}=S $
We know the identity $ {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}} $
Now substituting the value of $ b $ as $ 1 $ , then
 $ {{\left( a-1 \right)}^{3}}={{a}^{3}}-1-3{{a}^{2}}+3a $
Now multiplying minus $ \left( - \right) $ to the above equation, then
 $ -{{\left( a-1 \right)}^{3}}=1-3a+3{{a}^{2}}-{{a}^{3}} $
Now add the value $ {{a}^{3}} $ to the both side of above equation, then
 $ {{a}^{3}}-{{\left( a-1 \right)}^{3}}=3{{a}^{2}}-3a+1 $
Now substituting the values $ 1,2,3,4,...,n $ in $ a $ one by one, then
 $ \begin{align}
  & {{1}^{3}}-{{\left( 1-1 \right)}^{3}}=3{{\left( 1 \right)}^{2}}-3\left( 1 \right)+1 \\
 & {{2}^{3}}-{{\left( 2-1 \right)}^{3}}=3{{\left( 2 \right)}^{2}}-3\left( 2 \right)+1 \\
 & {{3}^{3}}-{{\left( 3-1 \right)}^{3}}=3{{\left( 3 \right)}^{2}}-3\left( 3 \right)+1 \\
 & {{4}^{3}}-{{\left( 4-1 \right)}^{3}}=3{{\left( 4 \right)}^{2}}-3\left( 4 \right)+1 \\
 & . \\
 & . \\
 & . \\
 & {{n}^{3}}-{{\left( n-1 \right)}^{3}}=3{{\left( n \right)}^{2}}-3\left( n \right)+1
\end{align} $
Now adding the all the equations then
 $ \begin{align}
  & {{1}^{3}}-{{\left( 1-1 \right)}^{3}}+{{2}^{3}}-{{\left( 2-1 \right)}^{3}}+{{3}^{3}}-{{\left( 3-1 \right)}^{3}}+...+{{n}^{3}}-{{\left( n-1 \right)}^{3}}=3{{\left( 1 \right)}^{2}}-3\left( 1 \right)+1+3{{\left( 2 \right)}^{2}}-3\left( 2 \right)+1+3{{\left( 3 \right)}^{2}}-3\left( 3 \right)+1+...+3{{\left( n \right)}^{2}}-3\left( n \right)+1 \\
 & {{n}^{3}}-{{0}^{3}}=3\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{n}^{2}} \right)-3\left( 1+2+3+4+...+n \right)+\left( 1+1+1+1+....n\text{ times} \right)
\end{align} $
Now substituting the value $ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...+{{n}^{2}}=S $ in the above equation and the value $ 1+2+3+4+...+n=\dfrac{n\left( n+1 \right)}{2} $ , then
 $ \begin{align}
  & {{n}^{3}}=3\left( S \right)-3\left( \dfrac{n\left( n+1 \right)}{2} \right)+n \\
 & 3S={{n}^{3}}+\dfrac{3}{2}n\left( n+1 \right)-n \\
 & =n\left( {{n}^{2}}-1 \right)+\dfrac{3}{2}n\left( n+1 \right)
\end{align} $
Using the formula $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ in the above equation, then
 $ \begin{align}
  & 3S=n\left( n+1 \right)\left( n-1 \right)+\dfrac{3}{2}n\left( n+1 \right) \\
 & =n\left( n+1 \right)\left[ n-1+\dfrac{3}{2} \right] \\
 & =n\left( n+1 \right)\left( \dfrac{2n-2+3}{2} \right) \\
 & S=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}
\end{align} $
So we have the value of sum of squares of the $ n $ natural numbers, now we need to find the average of the squares of the $ n $ natural numbers by using the formula
Average of square of $ n $ natural numbers $ =\dfrac{\text{Sum of squares of }n\text{ natural numbers}}{n} $
                                                                           $ \begin{align}
  & =\dfrac{S}{n} \\
 & =\dfrac{\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}{n} \\
 & =\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6n} \\
 & =\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}
\end{align} $
Hence the required value is $ \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6} $

Note: We can also find the solution for this problem in another method. We can manually find the average of the squares of first $ 5 $ numbers and check the result by substituting $ n=5 $ in the given options. After that calculate the average of first $ 10 $ numbers and check the result by substituting $ n=10 $ in the given options.