10-seconds challenge-7

Algebra Level 4

x 100 1 + x + x 2 + x 3 + + x 200 \large \dfrac{x^{100}}{1+x+x^{2}+x^{3}+ \cdots +x^{200}}

For x > 0 x>0 , the maximum value of the expression above is 1 N \dfrac{1}{N} for some integer N N . Find N N .


This is a part of 10-seconds challenge .


The answer is 201.

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1 solution

Rishabh Jain
Mar 7, 2016

Expression can be written as : 1 1 x 100 + 1 x 99 + x 99 + x 100 \dfrac{1}{\color{#0C6AC7}{\frac{1}{x^{100}}}+\color{#D61F06}{\frac{1}{x^{99}}}+\cdots \color{#D61F06}{x^{99}}+\color{#0C6AC7}{x^{100}}} Applying A M G M AM\geq GM to denominator , we can see minimum value of denominator is 201 201 . Hence maximum value of expression is 1 201 \boxed{\large {\color{#20A900}{\dfrac{1}{201}}}} , N = 201 \therefore N=201 . Equality occurs when x = 1 x=1 .

Nice problem and solution!

Anik Mandal - 5 years, 3 months ago

Easy problem and nice solution......same way!!!

abc xyz - 5 years, 3 months ago

Applying AM ≥ GM means?

Suneel Kumar - 5 years, 3 months ago

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See this

Rishabh Jain - 5 years, 3 months ago

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