Working through an upper-school pre-calculus book which starts with a bit of revision. Problem is that one of the questions in the development section of exercises is solvable using a technique they’ve not yet taught called ‘telescoping’.
If I process the equation by hand, I see the telescoping mechanism in action: all middle terms cancel, leaving only the first and last. However, the answer in the back of the book as well as the answer my CAS calculator generates has the opposite polarity to my own (theirs is positive where as mine is negative).
Had a look at some examples of telescoping online but I’m still confused as to why my handworked example is wrong.
The calculation is:
$$\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\ldots +\dfrac{1}{\sqrt{15}+\sqrt{16}}$$
The correct answer is $3$, but when I work it out I get $-3$.
I begin by rationalising the denominators by multiplying each term (both the numerator and the denominator) by the denominator with the operator inverted, such that the difference of squares is formed in the denominator, cancelling the radical and the terms in all of the denominators equal one, like so:
$$\dfrac{\sqrt{1}-\sqrt{2}}{\left( \sqrt{1}\right) ^{2}-\left( \sqrt{2}\right) ^{2}}+\dfrac{\sqrt{2}-\sqrt{3}}{\left( \sqrt{2}\right) ^{2}-\left( \sqrt{3}\right) ^{2}}+\dfrac{\sqrt{3}-\sqrt{4}}{\left( \sqrt{3}\right) ^{2}-\left( \sqrt{4}\right) ^{2}}+\ldots +\dfrac{\sqrt{15}-\sqrt{16}}{\left( \sqrt{15}\right) ^{2}-\left( \sqrt{16}\right) ^{2}}$$
Simplified:
$$\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+\ldots +\dfrac{\sqrt{15}-\sqrt{16}}{15-16}$$
The penultimate operation looks like this:
$$\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+\ldots \sqrt{15}-\sqrt{16}$$
So all of the middle terms are cancelling (telescoping) except the first and last, leaving:
$$\sqrt{1}-\sqrt{16}=1-4=-3$$
Clearly my reasoning is flawed. Where does my error(s) creep in?
$$\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{15}-\sqrt{16}}{15-16}$$ Your work till here is correct. Let's evaluate each of this in turn. We notice that the deominantor of all the fractions is $-1$, so they evaluate to the following: $$-(\sqrt{1}-\sqrt{2})-(\sqrt{2}-\sqrt{3})-(\sqrt{3}-\sqrt{4})-...-(\sqrt{15}-\sqrt{16}) =\\ \sqrt{2} - \sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{16}-\sqrt{15}$$ Since all the terms ranging from $\sqrt{2}$ to $\sqrt{15}$ cancel out, the answer is $$\sqrt{16}-\sqrt{1} = 4-1 = \boxed{3}.$$
I think you forgot the $-1$ when expanding the terms, and that's why you got $-3$ as the final answer.