Mathxyz - upsc.xyz
0 votes

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?

asked Aug 17, 2017 by randomisation (1,920 points)  

1 Answer

0 votes
Best answer

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 randomisation (1,920 points)  
selected Aug 19, 2017 by randomisation
Welcome to MathXyz, where you can ask questions and receive answers from other members of the community. Please strictly ask questions from UPSC Mathematics syllabus.
37 questions
27 answers
2 comments
34 users