anonymous
  • anonymous
What is the sum of the first n terms of this series? 4 + 6 + 8 + 10 + 12 + ⋯
Mathematics
jamiebookeater
  • jamiebookeater
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geerky42
  • geerky42
Hmm. Would you know the answer if sum is 1+2+3+4+... instead?
anonymous
  • anonymous
Idk
zzr0ck3r
  • zzr0ck3r
I would do what @geerky42 said \(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\) Multiply every term in that \(\uparrow\) equation by \(4\)

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geerky42
  • geerky42
4?
geerky42
  • geerky42
I would say multiply that by 2.
geerky42
  • geerky42
Be careful here. term "2" is missing!
geerky42
  • geerky42
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+...
anonymous
  • anonymous
Can u explain little bit further I'm still lost
geerky42
  • geerky42
Do you get what @zzr0ck3r said?
anonymous
  • anonymous
Not really
geerky42
  • geerky42
You don't know the formula 1+2+3+...+n = \(\dfrac{n(n+1)}{2}\)?
anonymous
  • anonymous
No these problems are ones I have trouble with
geerky42
  • geerky42
oh well, let's just stick to fact that \(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\) For now.
geerky42
  • geerky42
OR do you want me to explain why this is true?
anonymous
  • anonymous
Yes please
zzr0ck3r
  • zzr0ck3r
Sorry I thought it was 4,8,12,...

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