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PRATIK subramaniyam SAXENA Grade: 9
        

Evaluate-


          summation r = 1 to r =n , r/[4(r^2)+1],when n tends to infinite.

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

summation r/[4r2 + 1] limit n-> infinity


summation (r) n(n+1)/2                  ( first n natural numbers)


summation (r2) n(n+1)(2n+1)/6     (sum of squares of first n nutural numbers)


so , summation r = 1 to r =n , r/[4(r^2)+1]


               = 3(n(n+1))/2[2(n(n+1)(2n+1) + 3]   lim n-> infinity


 put , n = 1/y


                =3(y+1)/2[2(y+1)(y+2) + 3y2]   limit y->0


now on taking limit


               = 3(1)/2[4] = 3/8


so , the value is 3/8 ...


approve if u like my ans...


 



6 years ago
PRATIK subramaniyam SAXENA
30 Points
										

SORRY !!!!!!!! The answer is 1/4


secondly u have done wrong, as i know we cant put the direct values of sum of n terms in the denominator .


we make a series and then try to solve it. but i did like that but it became very tough.

6 years ago
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