Induction question

Question

I have to prove that $$P(n):\quad 1^2-2^2+3^2-\dots+(-1)^{n+1}n^2=(-1)^{n+1}T_n$$ where $T_n=1+2+\ldots+n=\frac{n(n+1)}{2}$.

I know I have to solve by induction.

So, I showed a base case that when $n=1$, then $P(1)$ is true.

And then I did
$$P(n+1):\quad (-1)^{n+2}(n+1)^2+(-1)^{n+1}\frac{n(n+1)}{2}$$

but then I guess I'm messing up on my arithmetic part, because I can't get it to equal $P(n+1)$.

Answer 1

You’re on the right track.

$$\begin{align*} (-1)^{n+2}(n+1)^2+(-1)^{n+1}\frac{n(n+1)}2&=(-1)^{n+2}(n+1)^2-(-1)^{n+2}\frac{n(n+1)}2\\ &=(-1)^{n+2}\left((n+1)^2-\frac{n(n+1)}2\right)\\ &=(-1)^{n+1}\cdot\frac{2(n+1)^2-n(n+1)}2\;; \end{align*}$$

can you simplify the numerator to finish it from here?

Answer 2

Factor out $(-1)^{n+2}(n+1)$. The quotient is $$(n+1)-\frac n2=\frac{n+2}2$$ since $(-1)^{n+2}$ and $(-1)^{n+1}$ have opposite signs.