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suneja

  • 3 years ago

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

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  1. integralsabiti
    • 3 years ago
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    |dw:1350388976936:dw|

  2. suneja
    • 3 years ago
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    how did u get tis

  3. mahmit2012
    • 3 years ago
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    |dw:1350389094272:dw|

  4. integralsabiti
    • 3 years ago
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    difference of squares on denominator a^2-b^2=(a-b)(a+b)

  5. suneja
    • 3 years ago
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    dats partial sum rite

  6. mahmit2012
    • 3 years ago
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    yes

  7. suneja
    • 3 years ago
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    @mahmit2012 but ans is 10/21 can u show ur steps of partial sums

  8. Coolsector
    • 3 years ago
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    |dw:1350390395436:dw|

  9. Coolsector
    • 3 years ago
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    because 2n when n = 10 is 20 .. just to do the correction for mahmit2012

  10. suneja
    • 3 years ago
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    @Coolsector thanks..

  11. mahmit2012
    • 3 years ago
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    oh yes uper bound is 10 .

  12. suneja
    • 3 years ago
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    ya got it finally:) thanks ppl

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