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?
anyone? please, help would really come in handy, I don't know where to go from this.
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.
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?
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.
oh, okay! yeah, I really get it now; thanks so much for taking the time to explain it. :)
Sure! Let's break down the problem step by step.
To show that the sum of the squares of any five consecutive integers is divisible by 5, we can start by assuming the first consecutive integer as "n".
The five consecutive integers would be: n, (n+1), (n+2), (n+3), and (n+4).
Now, let's find the sum of their squares:
Sum of squares = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2
Expanding each square term:
Sum of squares = n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) + (n^2 + 6n + 9) + (n^2 + 8n + 16)
Simplifying:
Sum of squares = 5n^2 + 20n + 30
Now, we need to show that the sum of squares is divisible by 5. To do this, we can factor out 5 from the expression:
Sum of squares = 5(n^2 + 4n + 6)
Since (n^2 + 4n + 6) is an integer (as the sum of consecutive integers is always an integer), multiplying it by 5 makes the entire expression divisible by 5.
Therefore, we have shown that the sum of the squares of any five consecutive integers is indeed divisible by 5.