Question

If $f\left(x\right)=\frac{k^x}{k^x+\sqrt{k}}(k>0)$ then ${\sum }_{r=1}^{2n-1}2f\left(\frac{r}{2n}\right)=an+b$ where $a-b$ is equal to

Answer 1

$f\left(x\right)=\frac{k^x}{(k^x+\sqrt{k})}$

$f\left(x\right)=\frac{k^x}{(k^x+k^{\left(1/2\right)})}$ .......... $\left(i\right)$

On Dividing eq(i) by $k^x$, we get

$f\left(x\right)=\frac{1}{1+k^{(1/2-x)}}$ ........ $\left(ii\right)$

As $\frac{a^b}{a^c}=a^{(b-c)}$

On putting $x=\frac{r}{2n}$ in eq(ii), we get

$f\left(x\right)=\frac{1}{1+k^{(\frac{1}{2}-\frac{r}{2n})}}$

$f\left(x\right)=\frac{1}{1+k^{\left(\frac{n-r}{2n}\right)}}$

Now,

Let $A={\sum }_{r=1}^{2n-1}2f\left(\frac{r}{2n}\right)=2{\sum }_{r=1}^{2n-1}f\left(\frac{r}{2n}\right)$

Putting $r=1,........,(n-1),n,(n+1),.....,(2n-1)$, we get the following values respectively

$A=2\times [\frac{1}{1+k^{\frac{n-1}{2n}}}+......+\frac{1}{1+k^{\frac{1}{2n}}}+\frac{1}{2}+\frac{1}{1+k^{\frac{-1}{2n}}}+......+\frac{1}{1+k^{\frac{-(n-1)}{2n}}}]$

$A=2\times [\frac{1}{1+k^{\frac{n-1}{2n}}}+......+\frac{1}{1+k^{\frac{1}{2n}}}+\frac{1}{2}+\frac{k^{\frac{1}{2n}}}{k^{\frac{1}{2n}}+1}+......+\frac{k^{\frac{n-1}{2n}}}{k^{\frac{n-1}{2n}}+1}]$

On simplifying by adding first and last term; second and second last term ............; (n-1)th term and (n+1)th term

We get,

$A=2\times [1+1+1+1+....(n-1)times+\frac{1}{2}]$

$A=2\times [n-1+\frac{1}{2}]$

$A=2\times [n-\frac{1}{2}]$

$A=2n-1$ ------eq(iii)

According to question, $A=an+b$;

So, on comparing it with eq(iii) we can say that

$a=2;$ and $b=-1;$

So, $a-b=2+1=3$

$a-b=3$