1wel manipal thts a standard method hehehe [3][3][3]
This is a repost. The last time I posted this one it got buried somewhere
Find \sum_{r=1}^n \tan ^{-1} \frac{2r}{2r^4+1}
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19 Answers
341lets request nishant sir to colour it deep red coz tapan is blushing!!
21THIS IS A GR8 DAY FOR ME :
GETTING PRAISED FRM "PROPHET SIR" is a real honour!! [1]
thank you Sir
341wow! that was real fast. I almost spent a whole night wondering what to do! Good Job bro
21MY MAETHOD'S MAIN STAY :
divide n multiply 2
then it can b expressed difference of 2 terms, namely
2r2 + 1 + 2r && 2r2 + 1 - 2r
then it gets simpler
62I think this question was the first time i realised how good Prophet sir actually is :)
The proof i remember was so over the top (of my head) that i realised that I wud not have able to think the logic myself. :)
btw this is not even close to his best ;)
341deepansh - answer is close. maybe there is a typo in your post
11SIR YEH TO KUCHH IIT WAALE QUES JAISE LAG RAHA RAHA HAI
ITS A VERY GOOD 1
SO LETS THINK[12][12][12]
1wel i think we shud convert it into d form tan-1A-tan-1B.....and all will be cancelled except sum terms and it will be easy to calculate........BUT IM NOT GETTIN IT
341For doing that you must be able to convert the summand to the form
\lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{n} f\left(\frac{k}{n}\right)
See if that is possible here
11@anirudh that can be done only when n→∞
this cannot be used here
here i think separate the term inside tan-1
11Nice opportunity to ask something......how to convert summation to integral??
Do the 1 and n in this question become the lower and upper limits??
and does the tan-1.... become the function to integrate??