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bahrom7893

  • 4 years ago

Need serious series help again..

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  1. bahrom7893
    • 4 years ago
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    |dw:1330836048396:dw|

  2. bahrom7893
    • 4 years ago
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    Show that: |dw:1330836111973:dw|

  3. bahrom7893
    • 4 years ago
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    @Mertsj can u help?

  4. mathg8
    • 4 years ago
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    did you plugged in few values ... nice series !

  5. bahrom7893
    • 4 years ago
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    ohh yea.. hold on, tnx

  6. Zarkon
    • 4 years ago
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    easy to prove by induction

  7. bahrom7893
    • 4 years ago
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    zarkon, more hints? i've solved like 40 series problems today, my brains are malfunctioning.

  8. Zarkon
    • 4 years ago
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    use induction...prove for n=1...assume true of n...show true for n+1

  9. Zarkon
    • 4 years ago
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    you can also do a direct derivation...

  10. Zarkon
    • 4 years ago
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    \[\sum _{k=1}^n \text{ln}\left[\frac{k(k+2)}{(k+1)^2}\right]\] \[=\text{ln}\left[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}\right]\] then show \[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}=\frac{n+2}{2(n+1)}\]

  11. Zarkon
    • 4 years ago
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    \[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}\] \[=\frac{(1\cdot2\cdot3\cdot4\cdots n)(3\cdot4\cdot5\cdots n\cdot (n+1)\cdot(n+2)}{2^2\cdot3^2\cdot4^2\cdots n^2\cdot(n+1)^2}\]

  12. Zarkon
    • 4 years ago
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    \[=\frac{(1\cdot2\cdot\cancel{3}\cdot\cancel{4}\cdots \cancel{n})(\cancel{3}\cdot\cancel{4}\cdot\cancel{5}\cdots \cancel{n}\cdot (n+1)\cdot(n+2)}{2^2\cdot\cancel{3^2}\cdot\cancel{4^2}\cdots \cancel{n^2}\cdot(n+1)^2}\] \[=\frac{2(n+1)(n+2)}{2^2(n+1)^2}=\frac{n+2}{2(n+1)}\]

  13. Zarkon
    • 4 years ago
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    see what I did?

  14. bahrom7893
    • 4 years ago
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    yes, TY!

  15. Zarkon
    • 4 years ago
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    yw

  16. Zarkon
    • 4 years ago
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    fun problem

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