$$ -\sqrt3 +\sqrt4 -\sqrt5+\sqrt6 $$
Stuck on this question.. I thought it would have something to do with $i$, but that isn't the case.
My thought was something along the line of: $$ -\sqrt3 +\sqrt4 -\sqrt5+\sqrt6 \;=\;\sum_{n=3}^5 (\sqrt n )(i^n)$$
On
I believe you are after $$-\sqrt{3}+\sqrt4-\sqrt5+\sqrt6$$ Suppose instead it was: $$\sqrt3+\sqrt4+\sqrt5+\sqrt6$$ Then in summation notation we would write this as $$\sum_{n=3}^{6}{\sqrt n}$$ To deal with the sign flipping, we use $(-1)^n$, because $(-1)^n=1$ if $n$ is even, and $(-1)^n=-1$ if $n$ is odd. Hence our series becomes $$\sum_{n=3}^{6}{(-1)^n\sqrt n}$$
$$-\sqrt3+\sqrt4 + -\sqrt5 +\sqrt6 = \sum_{n=3}^6 (-1)^n(\sqrt n )$$
works better.