Question

The mean of the numbers $1,2,3,4,.........n$ with the respective weights ${1^2} + 1,{2^2} + 2,........{n^2} + n$ is:
A. $\dfrac{{3n(n + 1)}}{{2(2n + 1)}}$
B. $\dfrac{{n(n + 1)}}{{2(2n - 1)}}$
C. $\dfrac{{3n + 1}}{4}$
D. $\dfrac{{3n + 1}}{3}$

Answer 1

Hint:
Here the weight represents the frequency. Let it represent by $w$
So let ${x_i} = i$ where $i = 1,2,3......n$ and ${w_i} = i + {i^2}$
This satisfies the given equation
So mean$ = \dfrac{{\sum {{w_i}{x_i}} }}{{\sum {{w_i}} }} = \dfrac{{\sum\limits_{i = 1}^n {(i + {i^2})i} }}{{\sum\limits_{i = 1}^n {(i + {i^2})} }}$

Complete step by step solution:
Here in this question we need to find the mean of the given numbers $1,2,3,4,.....n$ with their respective weights ${1^2} + 1,{2^2} + 2,........{n^2} + n$
This means that here the respective weights represent the frequency of the given numbers. Let us denote it by ${w_i} = i + {i^2}$
So we can draw the table as

Numbers ${x_i}$	$1$	$2$	$3$	……………	$n$
Weight ${w_i}$	${1^2} + 1$	${2^2} + 2$	${3^2} + 3$	$...........$	${n^2} + n$

So basically here we have ${x_i}$ as the numbers so here ${x_i} = i$
And ${w_i} = i + {i^2}$ here represents the weights of the given numbers.
As we know that mean$ = \dfrac{{\sum {{w_i}{x_i}} }}{{\sum {{w_i}} }} = \dfrac{{\sum\limits_{i = 1}^n {(i + {i^2})i} }}{{\sum\limits_{i = 1}^n {(i + {i^2})} }}$
$ = \dfrac{{\sum\limits_{i = 1}^n {(i + {i^2})i} }}{{\sum\limits_{i = 1}^n {(i + {i^2})} }}$
Mean$ = \dfrac{{\sum\limits_{i = 1}^n {{i^3} + \sum\limits_{i = 1}^n {{i^2}} } }}{{\sum\limits_{i = 1}^n {i + \sum\limits_{i = 1}^n {{i^2}} } }}$
As we know that $\sum\limits_{i = 1}^n i $$ = $$1 + 2 + 3 + ....... + n = \dfrac{{n(n + 1)}}{2}$
$\sum\limits_{i = 1}^n {{i^2}} = $${1^2} + {2^2} + {3^2} + ........{n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
$\sum\limits_{i = 1}^n {{i^3}} = $${1^3} + {2^3} + {3^3} + ........{n^3} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4}$
So we can write these values in the above equation we will get that
Mean$ = \dfrac{{({1^3} + {2^3} + {3^3} + ........{n^3}) + ({1^2} + {2^2} + {3^2} + ........{n^2})}}{{({1^2} + {2^2} + {3^2} + ........{n^2}) + (1 + 2 + 3 + ....... + n)}}$
$ = \dfrac{{\dfrac{{{n^2}{{(n + 1)}^2}}}{4} + \dfrac{{n(n + 1)(2n + 1)}}{6}}}{{\dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2}}}$
$ = \dfrac{{\dfrac{{n(n + 1)}}{2}\left( {\dfrac{{n(n + 1)}}{2} + \dfrac{{2n + 1}}{3}} \right)}}{{\dfrac{{n(n + 1)}}{2}\left( {1 + \dfrac{{2n + 1}}{3}} \right)}}$
So upon calculating we will get that
Mean$ = \dfrac{{3{n^2} + 3n + 4n + 2}}{{4n + 8}}$
$ = \dfrac{{3{n^2} + 7n + 2}}{{4(n + 2)}}$
$ = \dfrac{{3{n^2} + 6n + n + 2}}{{4(n + 2)}}$
$ = \dfrac{{3n(n + 2) + (n + 2)}}{{4(n + 2)}}$
$ = \dfrac{{(n + 2)(3n + 1)}}{{4(n + 2)}} = \dfrac{{3n + 1}}{4}$

Therefore we get that ${\text{mean}} = \dfrac{{3n + 1}}{4}$

Note:
We can use hit and trial method by putting $n = 3$ the we can find the mean of the $1,2,3$ of the respective weights ${1^2} + 1,{2^2} + 2,{3^2} + 3$
So we get that mean$ = \dfrac{{\sum {w.x} }}{{\sum w }} = \dfrac{{1(2) + 2(6) + 3(12)}}{{2 + 6 + 12}} = \dfrac{5}{2}$
Now we can put $n = 3$ in the given options and check that which option has the value $\dfrac{5}{2}$ when we put n as the value three. So that option will be the correct one.
A. $\dfrac{{3(3)(4)}}{{2(7)}} \ne \dfrac{5}{2}$
B. $\dfrac{{3(4)}}{{2(5)}} \ne \dfrac{5}{2}$
C. $\dfrac{{9 + 1}}{4} = \dfrac{5}{2}$
Hence option C is correct.