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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Expansion ..infinite series
Forum Index -> Differential Calculus like the article? email it to a friend.  
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mona840 (700)

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Given tat
 
1/12 + 1/22 + 1/32 + 1/42 + ....... = 2 / 6  &
1/12 + 1/32 + 1/52 +  ....... = 2 / 8
 
find de value of  
 
[ 0][ 1] log [(1+x ) / (1-x) ]  dx/x
 
 

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rhd92781 (686)

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is d ans (pi)^2/4

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log(1+x/1-x) = log(1+x)-log(1-x)
=(x-x2/2+x3/3-x^4/4 + ....) - (-x-x^2/2-x^3/3-x^4/4-.... infinity)
=2(x+x^3/3 + x^5/5 + .... upto infinity)
 
[ ][ ] log(1+x/1-x) dx/x = [0 ][ 1] 2(1+x^2/3+x^4/5 + ....)
=22/8
=2/4

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lokeshsardana (675)

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yup................rahul's answer is perfect.....

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mona840 (700)

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Thanx a lot Rahul
i got it
 
nw
if v change de ques to
 
 
[ 0][ 1] log ( 1- x ) dx/x
 
does de ans cums out to b   2/ 6
 
juss wanna chk ma ans

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rhd92781 (686)

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well in dat case goin d same way
ans vl be -(sum of two series)
=-(pi)^2/6-(pi)^2/8

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lokeshsardana (675)

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nup............................

answer will - (pi)^2/6

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mona840 (700)

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yep rite
 
4gt to tak tat -ve sign
 
it is cumin out to b - 2 / 6
 
thanx guys fr helpin me out !!

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