# If x>0, then find greatest value of the expression \[\dfrac{{{x}^{100}}}{1+x+{{x}^{2}}+{{x}^{3}}+.....+{{x}^{200}}}\].

Answer

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Hint:

Solve the expression, by using the AM-GM inequality. Consider non-negative n-natural numbers, form the expression of AM and GM,then solve it using the formula of n-natural numbers.

Complete step-by-step answer:

The inequality of Arithmetic mean (AM) and Geometric mean (GM), the AM-GM inequality states that ,the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list and further than that the two means are equal if and only if every number in the list is same.

i.e.\[AM\ge GM\]

Let us take the non-negative real numbers as,

\[1,x,{{x}^{2}},....,{{x}^{200}}\]

We know,\[AM\ge GM-(1)\]

AM is the summation of the digits by the number of digits,

\[AM=\dfrac{1+x+{{x}^{2}}+{{x}^{3}}+.....+{{x}^{200}}}{201}\]

GM is the product of the digits raised to the number of terms,

\[={{\left( 1.x.{{x}^{2}}.{{x}^{3}}.....{{x}^{200}} \right)}^{\dfrac{1}{201}}}\]

Substitute the values of AM and GM in equation (1).

\[\left( \dfrac{1+x+{{x}^{2}}+{{x}^{3}}+.....+{{x}^{200}}}{201} \right)\ge {{\left( 1.x.{{x}^{2}}.{{x}^{3}}.....{{x}^{200}} \right)}^{\dfrac{1}{201}}}\]

\[GM={{x}^{0}}.{{x}^{1}}.{{x}^{2}}.{{x}^{3}}.....{{x}^{200}}={{x}^{\left( 0+1+2+3+....+200 \right)}}\]

Formula of n-natural number in \[GM={{x}^{\left( \dfrac{200\times 201}{2} \right)}}\]

\[\begin{align}

& \Rightarrow \left( \dfrac{1+x+{{x}^{2}}+....+{{x}^{200}}}{201} \right)\ge {{x}^{\left( \dfrac{200\times 201}{2} \right)\left( \dfrac{1}{201} \right)}} \\

& \left( \dfrac{1+x+{{x}^{2}}+....+{{x}^{200}}}{201} \right)\ge {{x}^{100}} \\

\end{align}\]

Now cross multiplying them,

\[\dfrac{1}{201}\ge \dfrac{{{x}^{100}}}{1+x+{{x}^{2}}+....+{{x}^{200}}}\]

\[\therefore \]We got the greatest value of expression as\[\dfrac{1}{201}\].

Note:

By using AM-GM at the denominator which is greater than the total number of digits 201.

So this will become\[{{\left( 201\sum\limits_{i=0}^{200}{{{x}^{i}}} \right)}^{\dfrac{1}{201}}}\], thus you get denominator is always more than\[201.{{x}^{100}}\]. Thus the max value is \[\dfrac{1}{201}\]at 1.

Which also denotes that the expression is less than or equal to\[\dfrac{1}{201}\].

Solve the expression, by using the AM-GM inequality. Consider non-negative n-natural numbers, form the expression of AM and GM,then solve it using the formula of n-natural numbers.

Complete step-by-step answer:

The inequality of Arithmetic mean (AM) and Geometric mean (GM), the AM-GM inequality states that ,the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list and further than that the two means are equal if and only if every number in the list is same.

i.e.\[AM\ge GM\]

Let us take the non-negative real numbers as,

\[1,x,{{x}^{2}},....,{{x}^{200}}\]

We know,\[AM\ge GM-(1)\]

AM is the summation of the digits by the number of digits,

\[AM=\dfrac{1+x+{{x}^{2}}+{{x}^{3}}+.....+{{x}^{200}}}{201}\]

GM is the product of the digits raised to the number of terms,

\[={{\left( 1.x.{{x}^{2}}.{{x}^{3}}.....{{x}^{200}} \right)}^{\dfrac{1}{201}}}\]

Substitute the values of AM and GM in equation (1).

\[\left( \dfrac{1+x+{{x}^{2}}+{{x}^{3}}+.....+{{x}^{200}}}{201} \right)\ge {{\left( 1.x.{{x}^{2}}.{{x}^{3}}.....{{x}^{200}} \right)}^{\dfrac{1}{201}}}\]

\[GM={{x}^{0}}.{{x}^{1}}.{{x}^{2}}.{{x}^{3}}.....{{x}^{200}}={{x}^{\left( 0+1+2+3+....+200 \right)}}\]

Formula of n-natural number in \[GM={{x}^{\left( \dfrac{200\times 201}{2} \right)}}\]

\[\begin{align}

& \Rightarrow \left( \dfrac{1+x+{{x}^{2}}+....+{{x}^{200}}}{201} \right)\ge {{x}^{\left( \dfrac{200\times 201}{2} \right)\left( \dfrac{1}{201} \right)}} \\

& \left( \dfrac{1+x+{{x}^{2}}+....+{{x}^{200}}}{201} \right)\ge {{x}^{100}} \\

\end{align}\]

Now cross multiplying them,

\[\dfrac{1}{201}\ge \dfrac{{{x}^{100}}}{1+x+{{x}^{2}}+....+{{x}^{200}}}\]

\[\therefore \]We got the greatest value of expression as\[\dfrac{1}{201}\].

Note:

By using AM-GM at the denominator which is greater than the total number of digits 201.

So this will become\[{{\left( 201\sum\limits_{i=0}^{200}{{{x}^{i}}} \right)}^{\dfrac{1}{201}}}\], thus you get denominator is always more than\[201.{{x}^{100}}\]. Thus the max value is \[\dfrac{1}{201}\]at 1.

Which also denotes that the expression is less than or equal to\[\dfrac{1}{201}\].

Last updated date: 17th Sep 2023

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