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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!!

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.

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 :
$\displaystyle n+n+1+\dots+n+4=5n+10$
The text talks about the square of the mean.
So it's $\displaystyle (\frac{5n+10}{5})^2=(n+2)^2=\dots$
Now the other side.
The sum of the squares is $\displaystyle 5n^2+{\color{red}20}n+30$
Hence, since the text talks about the mean of the squares, it's $\displaystyle \frac{5n^2+20n+30}{5}=\dots$
Can you continue ? :)

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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!)

You've got nē=nē, which is true.
Hence, your affirmation is ok.
But it's better doing this way :
The square of the mean is :
$\displaystyle (n+2)^2=n^2+4n+4$
The mean of the squares is :
$\displaystyle \frac{5n^2+20n+30}{5}=n^2+4n+6$
And it's obvious that the difference is 2 :)
Why is it better this way ? Because you're not really supposed to know the result :)