Why is $\frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2 =\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2}$?

Question

Why is $\frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2 =\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2}$?

My math book says it has to do with the fact that $(-1)^{n+1}=(-1)^{n+3}=(-1)^{n+2}(-1)$

But it just doesn't click for me... Any help in understanding this is much appreciated!

Answer 1

That would be correct... replace $(-1)^{n+1} $ by $(-1)^{n+3} $, since they differ by a factor $(-1)^2=1$.

Answer 2

We can factor the term: $$\frac{(-1)^{n+1}(n+1)}{2}$$ to give us: $$\frac{(-1)^{n+1}(n+1)}{2} \left[n + (-2)(n+1) \right] $$ $$=\frac{(-1)^{n+1}(n+1)}{2} [-n-2]$$ $$=\frac{(-1)^{n+2}(n+1)(n+2)}{2}$$ which is the same as the books answer.

Answer 3

Factor out a common factor of, say, $\frac{(-1)^{n+1}(n+1)}{2}$.

Now we have $\frac{(-1)^{n+1}(n+1)}{2}(n + 2(-1)(n+1))$.

If we factor out another $(-1)$, we get $\frac{(-1)^{n+2}(n+1)}{2}(-n + 2(n+1))$.

Answer 4

write the left Hand side as $$\frac{-(-1)^{n+2}n(n+1)}{2}+\frac{2(-1)^{n+2}(n+1)^2}{2}=\frac{(n+1)(-1)^{n+2}(n+1-n)}{2}$$