Question

If \[{\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n}\] is
(1) \[ > {n^n}\]
(2) \[{n^n}\left( n \right)!\]
(3) \[{n^n}{\left( {n!} \right)^r}\] for all \[r\]
(4) none of these

Answer 1

Hint: This question is from the topic of arithmetic mean and geometric mean i.e., (A.M. and G.M.). In solving this question, we will first use the formula of A.M. and G.M. After that we will use the concept of relationship between A.M. and G.M. Then we will simplify it and get our required answer.
Formulas used:
(1) Arithmetic mean of n terms,
\[A.M. = \dfrac{{{a_1} + {a_2} + {a_3} + .... + {a_n}}}{n}\]
(2) Geometric mean of n terms,
\[G.M. = {\left( {{a_1} \cdot {a_z} \cdot {a_3} \cdot {\text{ }}....{\text{ }} \cdot {a_n}} \right)^{\dfrac{1}{n}}}\]
(3) Relation between A.M and G.M
\[ \Rightarrow A.M \geqslant G.M.\]

Complete step-by-step answer:
The given series is,
\[{\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n}\]
And we are asked to find out the sum or we can say the value of the given series.
Now, we know that
Arithmetic mean of n terms, \[A.M. = \dfrac{{{a_1} + {a_2} + {a_3} + .... + {a_n}}}{n}\]
Here, \[{a_1} = {1^r},{\text{ }}{a_2} = {2^r},{\text{ }}{a_3} = {3^r},{\text{ }}.....,{\text{ }}{a_n} = {n^r}\]
\[\therefore A.M. = \dfrac{{{1^r} + {2^r} + {3^r} + .... + {n^r}}}{n}{\text{ }} - - - \left( 1 \right)\]
Now in the same way use the formula for geometric mean,
We know that,
Geometric mean of n terms, \[G.M. = {\left( {{a_1} \cdot {a_z} \cdot {a_3} \cdot {\text{ }}....{\text{ }} \cdot {a_n}} \right)^{\dfrac{1}{n}}}\]
Here, \[{a_1} = {1^r},{\text{ }}{a_2} = {2^r},{\text{ }}{a_3} = {3^r},{\text{ }}.....,{\text{ }}{a_n} = {n^r}\]
\[\therefore G.M. = {\left( {{1^r} \cdot {2^r} \cdot {3^r} \cdot {\text{ }}.....{\text{ }} \cdot {n^r}} \right)^{\dfrac{1}{n}}}{\text{ }} - - - \left( 2 \right)\]
Now, by using the relation between A.M and G.M.
i.e., \[A.M \geqslant G.M.{\text{ }} - - - \left( 3 \right)\]
Substitute the value from \[\left( 1 \right)\] and \[\left( 2 \right)\] in \[\left( 3 \right)\] , we get
\[\dfrac{{{1^r} + {2^r} + {3^r} + .... + {n^r}}}{n} \geqslant {\left( {{1^r} \cdot {2^r} \cdot {3^r} \cdot {\text{ }}.....{\text{ }} \cdot {n^r}} \right)^{\dfrac{1}{n}}}\]
On multiplying both sides by \[n\] ,we get
\[ \Rightarrow \left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right) \geqslant n{\left( {{1^r} \cdot {2^r} \cdot {3^r} \cdot {\text{ }}.....{\text{ }} \cdot {n^r}} \right)^{\dfrac{1}{n}}}\]
Taking power \[n\] on both sides, we get
\[ \Rightarrow {\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} \geqslant {\left( {n{{\left( {{1^r} \cdot {2^r} \cdot {3^r} \cdot {\text{ }}.....{\text{ }} \cdot {n^r}} \right)}^{\dfrac{1}{n}}}} \right)^n}\]
\[ \Rightarrow {\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} \geqslant \left( {{n^n}{{\left( {{1^r} \cdot {2^r} \cdot {3^r} \cdot {\text{ }}.....{\text{ }} \cdot {n^r}} \right)}^{\dfrac{n}{n}}}} \right)\]
After simplifying, we get
\[ \Rightarrow {\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} \geqslant {n^n}\left( {{1^r} \cdot {2^r} \cdot {3^r} \cdot {\text{ }}.....{\text{ }} \cdot {n^r}} \right)\]
\[ \Rightarrow {\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} \geqslant {n^n}{\left( {1 \cdot 2 \cdot 3 \cdot {\text{ }}.....{\text{ }} \cdot n} \right)^r}{\text{ }} - - - \left( 4 \right)\]
We know that,
\[n! = 1 \cdot 2 \cdot 3 \cdot {\text{ }}.....{\text{ }} \cdot n\]
Therefore, equation \[\left( 4 \right)\] becomes,
\[{\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} \geqslant {n^n}{\left( {n!} \right)^r}\]
which can also be written as,
\[{\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} = {n^n}{\left( {n!} \right)^r}\]
And this result is true for all the values of \[r\]
Hence, the value of \[{\left( {{1^r} + {2^r} + {3^r} + ..... + {n^r}} \right)^n} = {n^n}{\left( {n!} \right)^r}\] for all \[r\]
So, the correct answer is “Option 3”.

Note: Whenever we have to solve these types of questions, always keep in mind that Arithmetic mean of n terms can never be less than their geometric mean.
Because we know that, the square of a number is always greater than or equals to \[0\] (if it’s real)
So, we can write \[{\left( {\sqrt a - \sqrt b } \right)^2} \geqslant 0\]
On expanding, we get
\[a + b - 2\sqrt {ab} \geqslant 0\]
\[ \Rightarrow a + b \geqslant 2\sqrt {ab} \]
which can also be written as,
\[ \Rightarrow \dfrac{{a + b}}{2} \geqslant \sqrt {ab} \]
And we know that Arithmetic mean of \[a\] and \[b\] is \[\dfrac{{a + b}}{2}\] and geometric mean of \[a\] and \[b\] is \[\sqrt {ab} \]
\[ \Rightarrow A.M \geqslant G.M\]
Hence, arithmetic mean can never be less than geometric mean.