# consecutive numbers

• Apr 27th 2008, 09:56 AM
Sweeties
consecutive numbers
hello everyone!

the square of the mean of the consecutive integers 1,2,3,4,5 is 9
the mean of the squares of the same 5 numbers is 11

show algebraically that the square of any five consecutive integers is always 2 less than the mean of the squares of those 5 consecutive integers

Here's what I got, but I have a sneaky feeling I have done something wrong cos I ended up with a quadratic!!
• Apr 27th 2008, 09:57 AM
Sweeties
I forgot to include the -2 on the right hand side in the image but I did include it in the working to get the final equation.
• Apr 27th 2008, 10:03 AM
Moo
Hello,

You also forgot that you had to compare the square of the mean, not the mean.

And you can't simplify the 5. I'll give you a (BIG :D) hint :

$n+n+1+\dots+n+4=5n+10$

The text talks about the square of the mean.

So it's $(\frac{5n+10}{5})^2=(n+2)^2=\dots$

Now the other side.
The sum of the squares is $5n^2+{\color{red}20}n+30$

Hence, since the text talks about the mean of the squares, it's $\frac{5n^2+20n+30}{5}=\dots$

Can you continue ? :)
• Apr 27th 2008, 01:31 PM
Sweeties
OK. Thanks for the reply. I had another go, I'm afraid it still doesn't look right!
I am enjoying it though!

OH, that should be n = n^2 at the end, sorry (again!)
• Apr 27th 2008, 01:38 PM
Moo
You've got nē=nē, which is true.
$(n+2)^2=n^2+4n+4$
$\frac{5n^2+20n+30}{5}=n^2+4n+6$