Question 914968
a. n = 1 --> *[tex \large (1+0)(1-0) = 1]
n = 10 --> *[tex \large (\sqrt{10} + 3)(\sqrt{10} - 3) = 1] (you can easily check this)


b. Difference of squares or FOIL:
*[tex \large (\sqrt{n} + \sqrt{n-1})(\sqrt{n} - \sqrt{n-1}) = (\sqrt{n})^2 - (\sqrt{n-1})^2 = n - (n-1) = 1]


c. Showing the statement holds in general is a much stronger proof - checking a few cases usually does not constitute a valid proof. In fact, checking a few cases is widely considered invalid unless you can show that you checked all possible cases, or all possible cases reduce to ones you've already checked.


For example, to show that n^6 leaves a remainder of 0 or 1 when divided by 7, you only need to check n = 0,1,2,...,6 (do you see why?). However this is true by Fermat's little theorem.