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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: A question on progressions..........
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irlmaks (125)

Cool goIITian

Olaaa!! Perrrfect answer. 21  [31 rates]

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total posts: 61    
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I think i have a better solution
 
tr+1-tr =(4r+3)(4r+7)(4r+11)(4r+15)-(4r-1)(4r+3)(4r+7)(4r+11)
         =16(4r+3)(4r+7)(4r+11)
         =16tr /4r-1
implies   19tr+tr+1=4r.tr+1- 4(r-1)tr
 
telescoping we get 4n.tn+1-4.0(t1)=20Sn+tn+1-t1
 
hence Sn=[(4n-1)(4n+3)(4n+7)(4n+11)(4n+15)+3.7.11.15]/20
            
DONT HESITATE TO ASK FOR ELOBERATION OF MY SOLUTION IF U THINK MY SOLUTION IS OVER ELAGENT

i may not be able to give the best solution!
But i can definetly give a good solution!
 this reply: 7 points  (with Olaaa!! Perrrfect answer.   in 2 votes )   [?]
 
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