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pythagoras123

  • 4 years ago

Calculation question: (see attachment)

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  1. pythagoras123
    • 4 years ago
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  2. Arnab09
    • 4 years ago
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    u know the answer? is it 209/4200 ??

  3. Arnab09
    • 4 years ago
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    whatever be the answer, u can proceed this way and get the answer: notice that all the terms are in the form: \[1/5(n-1)n(n+1) \] form so, u can write it this way: \[[(1/n-1)-(1/n+1)]/10n\] take the 1/10 common so, if a term is 1/r(r-1) - 1/r(r+1), the next term is 1/r(r+1) - 1/(r+1)(r+2) when adding these two, note that the term 1/r(r+1) gets cancelled so, proceeding this way, we can get 1/1*2 - 1/20*21 = 209/420 multiply with 1/10, u will get 209/4200

  4. pythagoras123
    • 4 years ago
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    But Arnab09, 1/5(n−1)n(n+1) does not quite tally with the terms for the equation. (n-1) for the first term is zero, and any number multiplied by zero remains zero...

  5. Arnab09
    • 4 years ago
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    no, here the terms start with n=2

  6. Arnab09
    • 4 years ago
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    i have taken the middle term as n.. :)

  7. Ishaan94
    • 4 years ago
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    Goodjob arnab

  8. Arnab09
    • 4 years ago
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    oh! ok, thanx

  9. pythagoras123
    • 4 years ago
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    Now I understand... :)

  10. Taufique
    • 4 years ago
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    given series is:: Sn=(summation form n=1 to n=19)(1/5)((1/2r)-(1/r)+1/2(r+1))...expand this equation and you will get : Sn=(1/5)*(1/4-1/2(n+1)(n+2))....put here n=19 and Sn=209/4200..ans

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