1.

Find the value of (Sigma_(r=1)^(n) 1/r)/(Sigma_(r=1)^(n) k/((2n-2k+1)(2n-k+1))).

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> <br/> `A=sum_(<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)^(n)k/((2n-2k+1)(2n-k+1))` <br/> `=sum_(k=1)^(n)((2n-k+1)-(2n-2k+1))/((2n-2k+1)(2n-k+1))` <br/> `sum_(k=1)^(n)1/(2n-2k+1)-sum_(k=1)^(n)1/(2n-k+1)` <br/> `=(1/1+1/3+1/5+..+1/(2n-1))-(1/(n+1)+1/(n+2)+..+1/(2n))` <br/> and `B=sum_(r=1)^(n)1/r=1/1+1/2+1/3+...+1/n+1/(n+1)+1/(n+2)+..+1/(2n))-(1/1+1/3+1/5+..+1/(2n-1))` <br/> `=1/2+1/4+1/6+...1/(2n)` <br/> `=1/2(1/1+1/2+1/3+..+1/n)` <br/> `=B/2` <br/> So, `B-A=1/2B` <br/> `rArrB/2=A` <br/> `rArrB/A=2`</body></html>


Discussion

No Comment Found

Related InterviewSolutions