- #1
MHB Proof by induction? No Idea what I should do :(
Click For Summary
The discussion focuses on proving a mathematical statement by induction, specifically the equation involving the sum of squares with alternating signs. The initial attempts to establish the general rule were corrected, emphasizing the need to simplify the right side using the known formula for the sum of the first n integers. The correct equation to prove is presented as the sum of squares equating to a specific expression involving k. The conversation concludes with guidance on how to manipulate the equation to complete the proof, stressing the importance of verifying the base case and the inductive step.
- #2
Opalg
Gold Member
MHB
- 2,778
- 13
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:
- #3
- 2,020
- 843
Actually, I think what you want isFriendlyCashew said:
[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
- #4
FriendlyCashew
- 2
- 0
Is this correct?
- #5
Opalg
Gold Member
MHB
- 2,778
- 13
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:
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?