How do you find the sum of the squares of consecutive even numbers?

Answer 1

What I would do is square each of the consecutive

even numbers, and then add their squares.

It depends on how complex you want the answer to be. If you need a formula to do it, then use the following.

If it's always starting at two, then use the formula:

Sum of even numbers' squares from 0 to w.

x=w/2

f(x) = (4*x^3+6*x^2+2*x)/3

If you put in 1, then you get the first even number squared. If you put in two, then you get the sum of the squares of the first two even numbers. Three will give you the sum of the squares of the first three even numbers.

If you need to vary where it starts (e.g. adding the squares of the even numbers from 8 to 26) the use that formula with the larger number (13, because 26 is the thirteenth even number) and then subtract the formula at the lower number minus one (3, since 8 is the fourth even number, and 4-1=3). F(13)=3276; F(3)=56; 3276-56=3220. So, the sum of the squares of the even numbers from 8 to 26 is 3220.

Sum of even numbers' squares from w to z.

x=(w/2)-1

y=z/2

f(y)-f(x)