Evaluate $\displaystyle \sqrt{2007\times2008\times2009\times2010-1}-2008^2$ without a calculator.
$\displaystyle n(n+1)(n+2)(n+3) = [n(n+3)][(n+1)(n+2)] $$\displaystyle = [n^2+3n][n^2 + 3n + 2] = [(n^2+3n+1)-1][(n^2+3n+1)+1]$$\displaystyle = (n^2 + 3n+1)^2 - 1$.
So are you sure it is $\displaystyle n(n+1)(n+2)(n+3) - 1$? If it was $\displaystyle n(n+1)(n+2)(n+3) + 1$ it would turn into a square. As the above squares?
Hello, math sucks!
Could there be a typo?
If the first minus-sign is a plus, there's a neat answer . . .
Let $\displaystyle x = 2008$Evaluate $\displaystyle \sqrt{2007\times2008\times2009\times2010 \:{\color{red}+}\:1}-2008^2$ without a calculator.
Under the radical, we have: .$\displaystyle (x-1)x(x+1)(x+2) +1$
. . . . . . $\displaystyle =\;[(x-1)(x+2)]\,[x(x+1)] + 1$
. . . . . . $\displaystyle = \;[x^2 + x - 2]\,[x^2 + x] + 1$
. . . . . . $\displaystyle =\;[x^2 + x - 1 - 1]\,[x^2+x-1 + 1] + 1$
. . . . . . $\displaystyle =\;\underbrace{[(x^2+x-1) - 1]\,[(x^2+x-1) + 1]}_{(a-b)(a+b)\:=\:a^2-b^2} + 1$
. . . . . . $\displaystyle =\;(x^2+x-1)^2 - 1^2 + 1$
. . . . . . $\displaystyle =\;(x^2+x-1)^2$
The problem becomes: .$\displaystyle \sqrt{x^2+x-1)^2} - x^2 \;=\;x^2+x-1-x^2 \;=\;x-1$
Therefore, the value is: .$\displaystyle 2008 - 1 \;=\;{\color{blue}2007}$
For some reason it seems my calculator ignores any constants I put under the square root. Ignore my last posts. There must be a problem with my calculator.
EDIT: Ah I think I know what it is. The product under the square root is a very large number and subtracting or adding 1 does not change it much. Therefore when finding the root, one gets the same answer.
I do not have to see anything. If $\displaystyle \sqrt{n}$ happens to be an integer then $\displaystyle \sqrt{n-2}$ shall not be an integer (where $\displaystyle n\geq 2$). So there is no way how you can get the same answer (which is an integer).
It is time to buy an new calculator.