Question

What is the mean of the squares of the first 20 natural numbers?
A. 15.5
B. 143.5
C. 65
D. 72

Answer 1

- Hint – We will solve this question by using the formula of the sum of the squares of the first $n$natural numbers and after that we will calculate the mean ( i.e., sum of the terms divided by the total number of the terms) of the squares of the first 20 natural numbers.

Complete step-by-step solution -

Now, the formula for the sum of the squares of the first $n$natural numbers is given by,
${S_n} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$………… (1)

The formula for ${S_n}$ found out using the following method:
${\left( {a + 1} \right)^3} - {a^3} = \left( {{a^3} + {1^3} + 3a\left( {a + 1} \right)} \right) - {a^3}$
$ \Rightarrow {\left( {a + 1} \right)^3} - {a^3} = 3{a^2} + 3a + 1$
Now, putting $a = 1,2,...,n$ , we get
$
  {\left( {1 + 1} \right)^3} - {1^3} = 3\left( {{1^2}} \right) + 3\left( 1 \right) + 1 \\
  {\left( {2 + 1} \right)^3} - {2^3} = 3\left( {{2^2}} \right) + 3\left( 2 \right) + 1 \\
  .... \\
  {\left( {n + 1} \right)^3} - {n^3} = 3{n^2} + 3n + 1 \\
 $

Adding, these the LHS neatly simplifies and the equation becomes
$
  {\left( {n + 1} \right)^3} - {1^3} = 3\left( {{1^2} + {2^2} + ..... + {n^2}} \right) + 3\left( {1 + 2 + ..... + n} \right) + n \\
   \Rightarrow \left( {{n^3} + 3{n^2} + 3n + 1} \right) - 1 = 3{S_n} + 3\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) + n \\
   \Rightarrow {n^3} + 3{n^2} + 3n - 3\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) - n = 3{S_n} \\
   \Rightarrow \dfrac{{2\left( {{n^3} + 3n + 3{n^2}} \right) - 3n\left( {n + 1} \right) - 2n}}{2} = 3{S_n} \\
   \Rightarrow 3{S_n} = \dfrac{{2{n^3} + \left( {6{n^2} - 3{n^2}} \right) + \left( {6n - 2n - 3n} \right)}}{2} \\
   \Rightarrow {S_n} = \dfrac{{2{n^3} + 3{n^2} + n}}{6} \\
   \Rightarrow {S_n} = \dfrac{{n\left( {2{n^2} + 3n + 1} \right)}}{6} \\
   \Rightarrow {S_n} = \dfrac{{n\left( {2{n^2} + 2n + n + 1} \right)}}{6} \\
   \Rightarrow {S_n} = \dfrac{{n\left( {2n\left( {n + 1} \right) + \left( {n + 1} \right)} \right)}}{6} \\
   \Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\
 $
This method can be extended to find the sum of the first $n$terms of natural number raised to a power.

Now, in this question, we have $n = 20$.
Putting $n = 20$in equation (1), we obtain
$
  {S_{20}} = \dfrac{{20\left( {20 + 1} \right)\left( {2\left( {20} \right) + 1} \right)}}{6} \\
   \Rightarrow {S_{20}} = \dfrac{{20\left( {21} \right)\left( {40 + 1} \right)}}{6} \\
   \Rightarrow {S_{20}} = \dfrac{{20 \times 21 \times 41}}{6} \\
 $

Now, we have to find out the meaning.
As we know that, Mean is equal to the sum of the terms divided by the total number of the terms.
Therefore,
Mean $ = \dfrac{{\dfrac{{20 \times 21 \times 41}}{6}}}{{20}}$
            $
   = \dfrac{{20 \times 21 \times 41}}{{6 \times 20}} \\
   = \dfrac{{287}}{2} \\
   = 143.5 \\
 $
Hence, the mean of the squares of the first 20 natural numbers is 143.5.
Thus, option B is the right answer.
Note – Natural numbers are a part of the number system which includes all the positive integers from 1 till infinity. These types of questions are very simple, but one should know both the formulas because this is the only method to solve this question.