How to deal with $(-1)^{k-1}$

Question

It's a problem on mathematical induction.

$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$

I have proved it for values of $n=1,2$.

Now I assume for $n=k$

$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.

$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$

I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.

Answer 1

You are making a mistake, in that you are assuming that $(-1)^k=1$.

You have \begin{align} (-1)^{k-1}\frac{k.(k+1)}{2}+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{(k+1)}{2} [k-(2k+2)]\\ \ \\ &=(-1)^{k-1}\frac{(k+1)}{2} [-(k+2)]\\ \ \\ &=(-1)^{k}\frac{(k+1)(k+2)}{2} \\ \ \\ \end{align}

Answer 2

L.H.S. of $P(k+1)$ should be $1^2-2^2+3^2-\cdots+(-1)^{k-1}k^2+(-1)^k(k+1)^2$

which is equal to

$\displaystyle \frac{(-1)^{k-1}k(k+1)}{2}+(-1)^k(k+1)^2=\frac{(-1)^{k}[-k(k+1)+2(k+1)^2]}{2}=\frac{(-1)^{k}(k+1)(k+2)}{2}$

Answer 3

If$$1^2-2^2+\cdots+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k(k+1)}2,$$then\begin{align}1^2-2^2+\cdots+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{k(k+1)}2+(-1)^k(k+1)^2\\&=(-1)^k\left(-\frac{k(k+1)}2+(k+1)^2\right)\\&=(-1)^k(k+1)\left(k+1-\frac k2\right)\\&=(-1)^k(k+1)\frac{k+2}2\\&=(-1)^k\frac{(k+1)(k+2)}2.\end{align}

Answer 4

you have to prove that $$(-1)^k\frac{k(k+1)}{2}+(-1)^k(k+1)^2=(-1)^k\frac{(k+1)(k+2)}{2}$$ or $$(-1)^{k-1}\frac{k(k+1)}{2}=(-1)^k\frac{(k+1)(k+2)}{2}-(-1)^k(k+1)^2$$ the right-hand side is given by $$(-1)^k(k+1)\left(\frac{(k+2)}{2}-k-1\right)$$ and this is equal to $$(-1)^k(k+1)\left(\frac{k+2-2k-2}{2}\right)$$ Can you finish?

Answer 5

You forgot a minus sign.

$\displaystyle P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(-1)^k(k+1)^2\\=(-1)^{k-1}\dfrac{k.(k+1)}{2}+(-1)^k(k+1)^2\\=\dfrac{(k+1)}{2} [(-1)^{k-1}.k +2(-1)^k (k+1)]\\=\dfrac{(k+1)}{2} [-(-1)^{k}.k +2(-1)^k (k+1)]\\=(-1)^k\dfrac{(k+1)}{2}[-k+2k+2]\\=(-1)^k\dfrac{(k+1)}{2}[k+2]\\=(-1)^k\dfrac{(k+1)(k+2)}{2}$

(Proved)