Question

$\sf \red{Given \: f(x) = \frac{9}{ {9}^{x} + 3}, find } \\ \rm\color{green}{f \bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{2}{1996} \bigg) + \dots f \bigg( \frac{1995}{1996} \bigg) }$​

Answer 1

Appropriate Question :-

Given that

$\rm \: f(x) = \dfrac{ {9}^{x} }{ {9}^{x} + 3}$

then find the value of

$\rm \: f\bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{2}{1996} \bigg) + \dots\dots + f\bigg( \frac{1995}{1996} \bigg)$

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = \dfrac{ {9}^{x} }{ {9}^{x} + 3}$

Now, let we consider

$\rm \: f(1 - x) = \dfrac{ {9}^{1 - x} }{ {9}^{1 - x} + 3}$

$\rm \: f(1 - x) = \dfrac{ \dfrac{9}{ {9}^{x} } }{ \dfrac{9}{ {9}^{x} } + 3}$

$\rm \: f(1 - x) = \dfrac{9}{9 + 3. {9}^{x} }$

$\rm \: f(1 - x) = \dfrac{3}{3 + {9}^{x} }$

So,

$\rm \: f(x) + f(1 - x) =\dfrac{ {9}^{x} }{3 + {9}^{x} } + \dfrac{3}{3 + {9}^{x} }$

$\rm \: f(x) + f(1 - x) =\dfrac{ {9}^{x} + 3}{3 + {9}^{x} }$

$\rm\implies \:\rm \: f(x) + f(1 - x) =1$

Now, Consider

$\rm \: f\bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{2}{1996} \bigg) + \dots f\bigg( \frac{1995}{1996} \bigg)$

$\rm = f\bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{2}{1996} \bigg) + f\bigg( \frac{3}{1996} \bigg) + \dots + f\bigg( \frac{998}{1996}\bigg) + \dots + + f\bigg( \frac{1994}{1996} \bigg)f\bigg( \frac{1995}{1996} \bigg)$

can be re-arranged as

$\rm = \bigg[f\bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{1995}{1996} \bigg)\bigg] + \bigg[f\bigg( \frac{2}{1996} \bigg)+ f\bigg( \frac{1994}{1996}\bigg)\bigg] \\ \rm \: \dots \: \dots+ \bigg[f\bigg( \frac{997}{1996} \bigg) + f\bigg( \frac{999}{1996} \bigg)\bigg] + f\bigg( \frac{1}{2} \bigg)$

can be further rewritten as

$\rm = \bigg[f\bigg( \frac{1}{1996} \bigg) + f \bigg (1 - \frac{1}{1996} \bigg)\bigg] + \bigg[f\bigg( \frac{2}{1996} \bigg)+ f\bigg(1 - \frac{2}{1996}\bigg)\bigg] \\ \rm \: \dots \: \dots+ \bigg[f\bigg( \frac{997}{1996} \bigg) + f\bigg(1 - \frac{997}{1996} \bigg)\bigg] + f\bigg( \frac{1}{2} \bigg)$

As

$\boxed{ \rm{ \:\:\rm \: f(x) + f(1 - x) =1 \: \: \: }}$

So, using this, we get

$\rm \: = \: \underbrace{1 + 1 + \dots + 1} + \dfrac{ \sqrt{9} }{ \sqrt{9} + 3 } \\ 997 \: times \: \: \: \: \: \: \: \: \: \: \: \:$

$\rm \: = \: 997 + \dfrac{3}{3 + 3}$

$\rm \: = \: 997 + \dfrac{3}{6}$

$\rm \: = \: 997 + \dfrac{1}{2}$

$\rm \: = \: 997 +0.5$

$\rm \: = \: 997.5$

Hence,

$\boxed{ \rm{ \:f\bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{2}{1996} \bigg) + \dots\dots + f\bigg( \frac{1995}{1996} \bigg) = 997.5 \: }}$

Answer 2

Given:-

$\: \sf {Given \: f(x) = \frac{9}{ {9}^{x} + 3}, find } \\ \rm{f \bigg( \frac{1}{1996} \bigg) + f \bigg ( \frac{2}{1996} \bigg) + \dots f \bigg( \frac{1995}{1996} \bigg) }$

Solution :-

$\sf \: = f(x) = \frac{ {9}^{x} }{ {9}^{x} + 3 } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ {9}^{1 - x} }{ {9}^{1 - x} + 3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{ \frac{9}{ {9}^{x} } }{ \frac{9}{ {9}^{x} } + 3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{9}{9 + 3. {9}^{x} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{3}{3 + {9}^{x} } \\ \sf \:$

So

$\sf \: = f(x) + f(x - 1) = \frac{3 + {9}^{x} }{3 + {9}^{x} } =1 \\ \: \begin{gathered}\rm \: = \:\underbrace{1 + 1 + \dots + 1} + \dfrac{ \sqrt{9} }{ \sqrt{9} + 3 } \\ 997 \: times \end{gathered} \\ \begin{gathered}\rm \: = \: 997 + \dfrac{3}{3 + 3} \\ \end{gathered} \\ \begin{gathered}\rm \: = \: 997 + \dfrac{3}{6} \\ \end{gathered} \\ \begin{gathered}\rm \: = \: 997 + \dfrac{1}{2} \\ \end{gathered} \\ \begin{gathered}\rm \: = \: 997 +0.5 \\ \end{gathered} \\ \begin{gathered}\rm \: = \: 997.5 \\\end{gathered}$

Answer :-

$\sf \fbox \red{ = 997.5}$