Question

If $x>0$, $\dfrac{{{x}^{n}}}{1+x+{{x}^{2}}+...+{{x}^{2n}}}$ is
A. $\le \dfrac{1}{2n+1}$
B. $<\dfrac{1}{2n+1}$
C. $\ge \dfrac{1}{2n+1}$
D. $>\dfrac{2}{2n+1}$

Answer 1

Hint: In this question, we are to find the given expression. By assuming a binomial expression $x+\dfrac{1}{x}\ge 2$, we can extract the required expression and its value. For this, the binomial expression is considered for $n$ terms, and by adding all of them we get an expression that is the reciprocal of the given expression. Here, we have took the binomial as greater than $2$, since we have given that $x>0$

Formula Used: If $x+\dfrac{1}{x}\ge 2$; then ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}\ge 2$ and so on.
Then, for the degree or the power $n$, we can write
${{x}^{n}}+\dfrac{1}{{{x}^{n}}}\ge 2$

Complete step by step solution: Given that, $x>0$. So, that we can write,
 $x+\dfrac{1}{x}\ge 2,{{x}^{2}}+\dfrac{1}{{{x}^{2}}}\ge 2,...,{{x}^{n}}+\dfrac{1}{{{x}^{n}}}\ge 2$
On adding all the terms, we get
$(x+\dfrac{1}{x})+({{x}^{2}}+\dfrac{1}{{{x}^{2}}})+...+({{x}^{n}}+\dfrac{1}{{{x}^{n}}})\ge 2n$
Rewriting the series as
$(x+{{x}^{2}}+...+{{x}^{n}})+(\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}+...+\dfrac{1}{{{x}^{n}}})\ge 2n$
Adding $1$ on both sides, we get
$1+(x+{{x}^{2}}+...+{{x}^{n}})+(\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}+...+\dfrac{1}{{{x}^{n}}})\ge 1+2n$
Then, on simplifying the above expression, we get
$\begin{align}
  & 1+(x+{{x}^{2}}+...+{{x}^{n}})+(\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}+...+\dfrac{1}{{{x}^{n}}})\ge 1+2n \\
 & (x+{{x}^{2}}+...+{{x}^{n}})+(1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}+...+\dfrac{1}{{{x}^{n}}})\ge 1+2n \\
\end{align}$
On dividing and multiplying the L.H.S by ${{x}^{n}}$, we get
$\begin{align}
  & \dfrac{{{x}^{n}}(x+{{x}^{2}}+...+{{x}^{n}})+{{x}^{n}}(1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}+...+\dfrac{1}{{{x}^{n}}})}{{{x}^{n}}}\ge 1+2n \\
 & \dfrac{({{x}^{n+1}}+{{x}^{n+2}}+...+{{x}^{2n}})+({{x}^{n}}+{{x}^{n-1}}+{{x}^{n-2}}+...+{{x}^{2}}+x+1)}{{{x}^{n}}}\ge 1+2n \\
 & \dfrac{1+x+{{x}^{2}}+...+{{x}^{2n}}}{{{x}^{n}}}\ge 1+2n \\
\end{align}$
Thus, the obtained expression is the reciprocal of the given expression. So, for writing it in reciprocal, according to the inequality rules, the inequality symbol $\ge $ is to be changed to $\le $.
That is,
$\dfrac{{{x}^{n}}}{1+x+{{x}^{2}}+...+{{x}^{2n}}}\le \dfrac{1}{1+2n}$
Therefore, the required condition is obtained.

Option ‘A’ is correct

Note: Here, the solution starts with an assumption that must satisfy the given condition for the variable. So, on simplifying the inequality we considered using normal operations like addition, subtraction, multiplication, and division. If our assumption satisfies the given condition, then on simplifying it we get the required expression.