find the sum of (1/(2^2-1) +1/( 4^2-1)+1/(6^2-1)+...+1/(20^2-1))

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find the sum of (1/(2^2-1) +1/( 4^2-1)+1/(6^2-1)+...+1/(20^2-1))

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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how did u get tis
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Other answers:

difference of squares on denominator a^2-b^2=(a-b)(a+b)
dats partial sum rite
yes
@mahmit2012 but ans is 10/21 can u show ur steps of partial sums
|dw:1350390395436:dw|
because 2n when n = 10 is 20 .. just to do the correction for mahmit2012
@Coolsector thanks..
oh yes uper bound is 10 .
ya got it finally:) thanks ppl

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