Question

Let the function $f:[0,1]\to R$be defined by $f\left(x\right)=\frac{4^x}{(4x+2)}$. Then the value of $f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\dots +f\left(\frac{39}{40}\right)–f\left(\frac{2}{2}\right)$is _____.

Answer 1

Evaluate the given expression:

The Given function, $f:[0,1]\to R$such that

$f\left(x\right)=\frac{4^x}{(4x+2)}...\left(i\right)$

Replacing $x$ by $1-x$

$$\begin{gathered} f(1-x)=\frac{4^{1-x}}{(4^{1-x}+2)} \\ =\frac{2}{(2+4x)}\dots \dots \left(ii\right) \end{gathered}$$

Adding the equation $\left(i\right)$ & $\left(ii\right)$

$\Rightarrow f\left(x\right)+f(1-x)=1\dots ..\left(iii\right)$

Now from , $\left(iii\right)$

$$\begin{gathered} \Rightarrow f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\dots +f\left(\frac{39}{40}\right)–f\left(\frac{1}{2}\right) \\ =f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+\dots \dots .+f\left(\frac{20}{40}\right)+\dots +f\left(1–\frac{3}{40}\right)+f\left(1–\frac{2}{40}\right)+f\left(1–\frac{1}{40}\right)–f\left(\frac{1}{2}\right) \\ =\left[f\left(\frac{1}{40}\right)+f\left(1–\frac{3}{40}\right)\right]+\left[f\left(\frac{2}{40}\right)+f\left(1–\frac{2}{40}\right)\right]+\left[f\left(\frac{3}{40}\right)+f\left(1–\frac{3}{40}\right)\right]\dots ..f\left(\frac{20}{40}\right)–f\left(\frac{1}{2}\right) \end{gathered}$$

Putting the value from equation $\left(iii\right)$,

$$\begin{gathered} =1+1+1+\dots +1(19times)+f\left(\frac{1}{2}\right)–f\left(\frac{1}{2}\right) \\ =19 \end{gathered}$$

Hence, the value of,

$f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\dots +f\left(\frac{39}{40}\right)–f\left(\frac{2}{2}\right)=19$