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rohitarura (76)

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This question was posted earlier also.But i could not understand the given solutions


Question:


Find the Sum upto n terms:



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jishnudas1991 (1020)

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Sum !!! Till what...........


The answer my friend
Is blowing in the wind.
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vineetbhakhar (2)

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see my frnd


th last term will be = 1.5.9.- - - - -(4n-3) / 3.7.11. - - - - -(4n-1)      using A.P.


so sequence becom 1/3 + 5/9 +  - -- - - - - - - + 4n-3/4n-1


it sum up to


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


 

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rohitarura (76)

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Find the Sum upto n terms
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rohitarura (76)

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How did u get the second step.
Please explain
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rohitarura (76)

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Anyone pleaaaaaaaaassssssssssseeeeeeeeeeeeee helppppppppppppp!!!!!!!!!!!!!!

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rohitarura (76)

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Kaymant Sir,Bhatt Sir,pleaseee........
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studen9t_iit (210)

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www.goiit.com/posts/list/15/algebra-easy-81260.htm


tell_me_which_part_have_you_not_understood,i'll_explain_it-to_you..


http://www.youtube.com/watch?v=0JurgT5GEnc
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rohitarura (76)

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I am sorry very much,but i could not understand the second half of the answer.Can u please,please make it simpler for me,please??
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jishnudas1991 (1020)

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Edited


 look at next post


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


 


The answer my friend
Is blowing in the wind.
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jishnudas1991 (1020)

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rohitarura


 


This is my solution.And  I don't know how vineetbhakhar simplified the general term to such an extent.



The answer my friend
Is blowing in the wind.
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jishnudas1991 (1020)

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Re:Pleasssseee, answer this question


The answer my friend
Is blowing in the wind.
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elessar_iitkgp (2385)

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Excellent work @
jishnudas1991 . Going through the method itself was a delight. Lovely piece of work.



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hsbhatt (5772)

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Fantastic work jishnu and this forum owes you for this. I had heard about this technique but I had never encountered it before. So, personally, thanks from me.


Time wounds all heels
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jishnudas1991 (1020)

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