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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 + ⋯
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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geerky42
 one year ago
Best ResponseYou've already chosen the best response.2Hmm. Would you know the answer if sum is 1+2+3+4+... instead?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I would do what @geerky42 said \(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\) Multiply every term in that \(\uparrow\) equation by \(4\)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2I would say multiply that by 2.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2Be careful here. term "2" is missing!

geerky42
 one year ago
Best ResponseYou've already chosen the best response.22(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
 one year ago
Best ResponseYou've already chosen the best response.0Can u explain little bit further I'm still lost

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2Do you get what @zzr0ck3r said?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2You don't know the formula 1+2+3+...+n = \(\dfrac{n(n+1)}{2}\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No these problems are ones I have trouble with

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2oh well, let's just stick to fact that \(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\) For now.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.2OR do you want me to explain why this is true?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I thought it was 4,8,12,...
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