Question

What is the sum of the first n terms of this series?

4 + 6 + 8 + 10 + 12 + ⋯

Answer 1

Hmm. Would you know the answer if sum is 1+2+3+4+... instead?

Answer 2

Idk

Answer 3

I would do what @geerky42 said

\(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\)

Multiply every term in that \(\uparrow\) equation by \(4\)

Answer 4

4?

Answer 5

I would say multiply that by 2.

Answer 6

Be careful here. term "2" is missing!

Answer 7

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+...

Answer 8

Can u explain little bit further I'm still lost

Answer 9

Do you get what @zzr0ck3r said?

Answer 10

Not really

Answer 11

You don't know the formula 1+2+3+...+n = \(\dfrac{n(n+1)}{2}\)?

Answer 12

No these problems are ones I have trouble with

Answer 13

oh well, let's just stick to fact that

\(1+2+3+4+...+n = \dfrac{n(n+1)}{2}\)

For now.

Answer 14

OR do you want me to explain why this is true?

Answer 15

Yes please

Answer 16

Sorry I thought it was 4,8,12,...