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anonymous

  • one year ago

What is the sum of the first n terms of this series? 4 + 6 + 8 + 10 + 12 + ⋯

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  1. geerky42
    • one year ago
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    Hmm. Would you know the answer if sum is 1+2+3+4+... instead?

  2. anonymous
    • one year ago
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    Idk

  3. zzr0ck3r
    • one year ago
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    I would do what @geerky42 said \(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\) Multiply every term in that \(\uparrow\) equation by \(4\)

  4. geerky42
    • one year ago
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    4?

  5. geerky42
    • one year ago
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    I would say multiply that by 2.

  6. geerky42
    • one year ago
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    Be careful here. term "2" is missing!

  7. geerky42
    • one year ago
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    2(1+2+3+4+...) = 2+4+6+8+... But since 2 is missing, you just subtract 2. 2+4+6+8+... - 2 = 4+6+8+...

  8. anonymous
    • one year ago
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    Can u explain little bit further I'm still lost

  9. geerky42
    • one year ago
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    Do you get what @zzr0ck3r said?

  10. anonymous
    • one year ago
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    Not really

  11. geerky42
    • one year ago
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    You don't know the formula 1+2+3+...+n = \(\dfrac{n(n+1)}{2}\)?

  12. anonymous
    • one year ago
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    No these problems are ones I have trouble with

  13. geerky42
    • one year ago
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    oh well, let's just stick to fact that \(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\) For now.

  14. geerky42
    • one year ago
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    OR do you want me to explain why this is true?

  15. anonymous
    • one year ago
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    Yes please

  16. zzr0ck3r
    • one year ago
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    Sorry I thought it was 4,8,12,...

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