If $f(x)-<!--\; -\; -->f(y)=\mathrm{I<!--\; I\; -->}n\left(\frac{x}{y}\right)+x-<!--\; -\; -->y$, then find $\underset{k=1}{\overset{20}{\sum <!--\; \sum \; -->}}{f}^{\prime }\left(\frac{1}{k^2}\right)$

The correct answer is 2890.

We are given the functional equation:

$$f(x)-f(y)=\ln \left(\frac{x}{y}\right)+x-y$$

By using the logarithmic property:

$$\ln \left(\frac{x}{y}\right)=\ln (x)-\ln (y)$$

We can rewrite the given equation as:

$$f(x)-f(y)=\ln (x)-\ln (y)+x-y$$

Rearranging the terms to group the terms of variable
$x$
on one side and the terms of variable
$y$
on the other side:

$$f(x)-\ln (x)-x=f(y)-\ln (y)-y$$

Since the left-hand side is a function of only
$x$
and the right-hand side is the same function of only
$y$
, and both variables are independent, this expression must be equal to a constant, say
$c$
:

$$f(x)-\ln (x)-x=c$$

Solving for
$f\left(x\right)$
:

$$f(x)=c+x+\ln (x)$$

Now, we differentiate
$f\left(x\right)$
with respect to
$x$
:

$$f^{\prime }(x)=1+\frac{1}{x}$$

To find the value of the term in the summation, we substitute
$x=\frac{1}{k^2}$
into the derivative expression:

$$f^{\prime }\left(\frac{1}{k^2}\right)=1+\frac{1}{\left(\frac{1}{k^2}\right)}=1+k^2$$

We are asked to calculate the sum from
$k=1$
to
$20$
:

$$\sum _{k=1}^{20}{f}^{\prime }\left(\frac{1}{k^2}\right)=\sum _{k=1}^{20}(1+k^2)$$

This sum can be split into two separate summations:

$$\sum _{k=1}^{20}1+\sum _{k=1}^{20}{k}^2$$

Calculating the first summation (sum of 1 repeated 20 times):

$$\sum _{k=1}^{20}1=20$$

Calculating the second summation (the sum of squares of the first 20 natural numbers) using the standard formula:

$$\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$$

For
$n=20$
:

$$\sum _{k=1}^{20}{k}^2=\frac{20\times 21\times 41}{6}=2870$$

Adding the two parts together gives the final result:

$$\text{Total Sum}=20+2870=2890$$