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.FriendlyCashew said: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 isFriendlyCashew said: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 $1$.FriendlyCashew said: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