Mathematical Induction proof with triangular number sequence w/ alternating positive and negative #'s

Question

Consider the following five equations:

1) 1 = 1

2) 1 – 4 = -(1 + 2)

3) 1 –4 + 9 = 1 + 2 + 3

4) 1 –4 + 9 –16 = -(1 + 2 + 3 + 4)

5) 1 –4 + 9 – 16 + 25 = 1 + 2 + 3 + 4 + 5

Conjecture the general formula suggested by these five equations, and prove your conjecture using induction.

This is what I got for the formula:

n(Sum)i=1 (−1)^i−1 * i^2 = -1/2 * (-1)^n *(n(n+1))

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Inductive Step: The formula should hold for n=k+1

k+1(Sum)i=1 (-1)^i+1 * i^2 = -1/2* (-1)^k+2 * (k+1)(k+2)

1) k+1(Sum)i=1 (-1)^i+1 * i^2

2) k(Sum)i=1 (-1)^i+1 * i^2 + k+1(Sum)i=1 (-1)^i+1 * i^2

3) -1/2*(-1)^k+2 * (k(k+1)) + (-1)^(k+1)+1 * (k+1)^2 _____By using IH

4) -1/2*[ (-1)^k+1 * (k(k+1)) + 2*(-1)^k+2 * 2*(k+1)^2 ]

I get to step 4 and know I have to factor out but I'm not sure how to go about pulling things out that will get me to where I need to be.

I really need some guidance :(

Thanks in advance

Answer 1

You are assuming that$$\sum_{k=1}^n(-1)^{k+1}k^2=(-1)^{n+1}\frac{n(n+1)}2.$$If this holds for some $n$, then\begin{align}\sum_{k=1}^{n+1}(-1)^{k+1}k^2&=\left(\sum_{k=1}^n(-1)^{k+1}k^2\right)+(-1)^{n+2}(n+1)^2\\&=(-1)^{n+1}\frac{n(n+1)}2+(-1)^{n+2}(n+1)^2\\&=(-1)^{n+1}\left(\frac{n(n+1)}2-(n+1)^2\right)\\&=(-1)^{n+1}(n+1)\left(\frac n2-(n+1)\right)\\&=(-1)^{n+1}(n+1)\left(-\frac n2-1\right)\\&=(-1)^{n+1}(n+1)\frac{-n-2}2\\&=(-1)^{n+2}\frac{(n+1)(n+2)}2.\end{align}

Answer 2

hint

You could use

$$i+(i+1)=(i+1)^2-i^2$$

$$3+4=4^2-3^2$$

and just prove that $$\sum_{i=1}^ni=\frac{n(n+1)}{2}.$$