Proof by induction? No Idea what I should do :(

[FriendlyCashew]

find the general rule and prove by induction

1 = 1
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 -16 = -(1 + 2 + 3 + 4)

I created this so far, but don't know if I am even going the correct direction

 

[Opalg]

find the general rule and prove by induction

1 = 1
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 -16 = -(1 + 2 + 3 + 4)

I created this so far, but don't know if I am even going the correct direction

View attachment 10459

You have the correct formula, except that $n^n$ should be $n^2$. I think your next step should be to simplify the right side of the formula by using the formula (which you probably know?) for the sum of the numbers $1$ to $n$. After that, you should be able to apply the method of induction.

 

[topsquark]

find the general rule and prove by induction

1 = 1
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 -16 = -(1 + 2 + 3 + 4)

I created this so far, but don't know if I am even going the correct direction

View attachment 10459

Actually, I think what you want is
[math]\sum_{n = 1}^k (-1)^{n - 1} n^2 = (-1)^{k + 1} \sum_{n = 1}^k n[/math]

Does this work for some value of k? (Sure, try k = 1.) So if it works for k, does it work for k + 1?

-Dan

 

[FriendlyCashew]

Is this correct?

 

[Opalg]

find the general rule and prove by induction

1 = 1
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 -16 = -(1 + 2 + 3 + 4)

I created this so far, but don't know if I am even going the correct direction

View attachment 10459

The equation that you are trying to prove by induction is $$\sum_{n = 1}^k (-1)^{n - 1} n^2 = (-1)^{k + 1} \sum_{n = 1}^k n$$. Your first attempt at this was wrong, and so was mine (sorry about that). But topsquark got it right. The next step is to use the fact that $$ \sum_{n = 1}^k n = \frac12k(k+1)$$. So you want to prove that $$\sum_{n = 1}^k (-1)^{n - 1} n^2 = \frac12(-1)^{k + 1}k(k+1)$$. That equation is true when $k=1$ because both sides are then equal to $1$.

Now suppose that the equation is true for $k$. You want to show that it is also true for $k+1$, namely that $$\sum_{n = 1}^{k+1} (-1)^{n - 1} n^2 = \frac12(-1)^{k + 2}(k+1)(k+2)$$. On the left side of the equation, that differs from the previous equation just by the addition of one extra term $(-1)^k(k+1)^2$. So starting with the known equation $$ \sum_{n = 1}^k (-1)^{n - 1} n^2 = \frac12(-1)^{k + 1}k(k+1)$$, add $(-1)^k(k+1)^2$ to each side to get $$\sum_{n = 1}^{k+1} (-1)^{n - 1} n^2 = \frac12(-1)^{k + 1}k(k+1) + (-1)^k(k+1)^2$$.

Now can you simplify the right side of that equation to complete the proof?