Let

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

Then, $f(100)$ is equal to?

Using partial fractions, $\frac{k}{(2n-2k+1)(2n-k+1)} = \frac{1}{2n-2k+1} - \frac{1}{2n-k+1}$

Let $f(n) = \frac{p}{q}$

Then $f(n+1) = \frac{p + \frac{1}{n+1}}{q - \frac{1}{2n+2} + \frac{1}{n+1}} \implies f(n+1) = \frac{p+\frac{1}{n+1}}{q + \frac{1}{2}\frac{1}{n+1}}$

$\implies f(n+1) = 2. \frac{p+\frac{1}{n+1}}{2q +\frac{1}{n+1}} ............ (1)$

Now,

$f(1) = \frac{\frac{1}{1}}{\frac{1}{1.2}} = 2$

Let $f(k) = 2 = \frac{p}{q} \implies p = 2q ........... (2)$

Then, using componendo and dividendo, and eqs (1) and (2), we get,
$f(k+1) = 2$

Hence, $f(100) = 2$

answered Aug 17, 2017 by (1,920 points)
selected Aug 19, 2017