Question

what is the formula used in finding the sum of consecutive even integers with an exponent of 2 each?

Answer 1

S(n) =n(n+1)(2n+1)/6

Answer 2

I don't entirely follow the question but..

n=2k is the even counting function, so I suppose you're asking for:
\[\large \sum_{k=1}^{\infty} (2k)^2\]

Answer 3

@agreene yup. that's what i'm asking for

Answer 4

S(n) =n(n+1)(2n+1)/6
this is the formula

Answer 5

S stands for sum :)

Answer 6

@myko i think that's the formula in finding the sum of the squares of CONSECUTIVE INTEGERS (not even)

Answer 7

ups, sry didn't read eaven. My bad

Answer 8

@myko that's alright :D

Answer 9

\[\sum_1^\infty (2k)^2=\sum_1^\infty(4k^2)=4\sum_1^\infty k^2\]From there, you get that the formula is just \[4\cdot {n(n+1)(2n+1) \over 6}\]

Answer 10

The sums I wrote should go to \(n\). Not \(\infty\).

Answer 11

@KingGeorge thank you so much! :D

Answer 12

You're welcome.