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 $P(n+1)$.

What you mean is that $P(n)$ is the statement that $$\sum_{k=1}^n(-1)^{k+1}k^2=(-1)^{n+1}T_n;,$$ and you want to prove that $P(n)$ is true for each positive integer $n$. — Brian M. Scott, Sep 21 '15 at 00:16

score 2 · Accepted Answer · answered Sep 21 '15 at 00:22

2

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?

answered Sep 21 '15 at 00:22

Brian M. Scott

616,228

Why is it that you subtract and not add the (-1) to get (-1)^(n+2) – ematth7 Sep 21 '15 at 00:33
@ematth7: I’m not sure what you mean. Changing the exponent by $1$ in either direction flips the sign, so $-(-1)^{n+2}=(-1)^{n+1}$, and that’s exactly what I used in the first step. – Brian M. Scott Sep 21 '15 at 00:36
Ohhh so $-(-1)^{n+1}=(-1)^{n+2}$ too? – ematth7 Sep 21 '15 at 00:38
@ematth7: Yes, exactly. – Brian M. Scott Sep 21 '15 at 00:39

score 1 · Answer 2 · answered Sep 21 '15 at 00:23

1

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.

answered Sep 21 '15 at 00:23

Bernard

175,478

Induction question

2 Answers2