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klimenkov

Prove for a medal. The sum of the first \(n\) terms of the geometric sequence \(S_n\). \(b_1\) is the first term, \(q\) is the ratio.

  • one year ago
  • one year ago

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  1. klimenkov
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    Please, write a formula for \(S_n\) and it's proof.

    • one year ago
  2. klimenkov
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    It is better, you don't use any sourses to help yourself.

    • one year ago
  3. helder_edwin
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    if \(b_n=b_1q^{n-1}\). Let \[ \large S_n=b_1+b_2++b_3\dots+b_n=b_1+b_1q+b_1q^2+\dots+b_1q^{n-1} \] then \[ \large qS_n=b_1q+b_1q^2+\dots+b_1q^n \]

    • one year ago
  4. experimentX
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    |dw:1349108015779:dw|

    • one year ago
  5. helder_edwin
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    from these \[ \large qS_n-S_n=b_1q^n-b_1 \] and if \(q\neq1\) \[ \large S_n=b_1\cdot\frac{q^n-1}{q-1} \]

    • one year ago
  6. klimenkov
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    So, who owns a medal? Who was really the first? I think @helder_edwin.

    • one year ago
  7. helder_edwin
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    i don't think that matters.

    • one year ago
  8. helder_edwin
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    i think the point is that if u found the posts useful.

    • one year ago
  9. klimenkov
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    Yes, you are right. Hope @experimentX won't be sad about this.

    • one year ago
  10. experimentX
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    doesn't matter much to me

    • one year ago
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