Question

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

4 + 6 + 8 + 10 + 12 + ⋯

Answer 1

answer is 8

Answer 2

common difference d=?

Answer 3

no its not @jot

Answer 4

why

Answer 5

\[\Sigma \ its and answer whith this

Answer 6

\[\Sigma \]

Answer 7

ok

Answer 8

@iGreen
@Kemist
@ParthKohli helppp

Answer 9

see??

Answer 10

@iGreen do u know this?

Answer 11

No, sorry..

Answer 12

;'O

Answer 13

@ParthKohli




do you know?

Answer 14

4 + 6 + 8 + 10 + 12 + ⋯
2*(2 + 3 + 4 + 5 + 6 + ⋯)

The odd fact you should memorize is this, 1 + 2 + 3 + ... + n = \(\dfrac{n(n+1)}{2}\)

Check out inside those parentheses. Do you see a relationship?

Answer 15

not really 0.0 :'0

Answer 16

1 + 2 + 3 + 4 + ... + n -- n terms starting with 1. Sum is \(\dfrac{n(n+1)}{2}\)

2 + 3 + 4 + ... + n + (n+1) -- n terms starting with 2. Sum is \(\dfrac{(n+1)(n+2)}{2} - 1\)

Final answer is what?

4 + 6 + 8 + 10 + 12 + ⋯
2*(2 + 3 + 4 + 5 + 6 + ⋯)

Answer 17

@geerky42 ??

Answer 18

@tkhunny I DONT KNOW D:<

Answer 19

\(2\left(\dfrac{(n+1)(n+2)}{2} + 1\right) = (n+1)(n+2) + 2\)