Question

Show that the sum of the squares of any five consecutive integers is divisible by 5.
I think I should do something with n+(n+1)+(n+2)+(n+3)+(n+4), but I have no idea where to go to from here. Could someone please help me?

Answer 1

anyone? please, help would really come in handy, I don't know where to go from this.

Answer 2

n+(n+1)+(n+2)+(n+3)+(n+4) is good.

Now, square each of those.
(n^2) + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2

(n+1)^2 = (n+1)(n+1) and use FOIL
n^2 + 2n + 1

Do the rest. (n+2)^2 = (n+2)(n+2) = n^2 + 4n + 4

(n+3)^2 = n^2 + 6n + 9

(n+4)^2 = n^2 + 8n + 16

Now put that back into our expression above.

(n^2) + (n^2 + 2n + 1) + (n^2 + 4n + 4) + (n^2 + 6n + 9) + (n^2 + 8n + 16)

Simplify.
5(n^2) + 20n + 30

You can now easily see that each term is divisible by 5.

(n^2) + 4n + 6 is the sum divided by 5.

Answer 3

sorry, but I'm a tad confused; I get FOIL, but I don't get how you got "(n+1)^2 = (n+1)(n+1) and use FOIL
n^2 + 2n + 1" from just the original equation. Sorry, but would you mind explaining it?

Answer 4

Not a problem.

For example, if you have 4^2, that is the same as 4*4. If you have x^2, that is the same as x*x. Squaring is just multiplying the number or quantity by itself.

(n+1)^2 = (n+1)(n+1)

Now, we are going to FOIL (n+1)(n+1).
First, outer, inner, last.
(n)(n) + (n)(1) + (1)(n) + (1)(1)
n^2 + n + n + 1
n^2 + 2n + 1

Do you understand it better? Maybe it was difficult to understand because we started with "(n^2) + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2." From that, I was just breaking down each part of it. In this case, it was the "(n+1)^2" part.

Answer 5

oh, okay! yeah, I really get it now; thanks so much for taking the time to explain it. :)