# Math Help - Proof by induction.

1. ## Proof by induction.

I tried this last night and i'm looking at it this mornng. I'm still stuck on a bit of it:

Observe that:

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

Guess the general law and prove it by induction.
My method is like this:

$1-2^2+3^2-4^2+.........-(n-1)^2+n^2=(-1)^{n+1}(\frac{n}{2})(2+(n-1))$

$1-2^2+3^2-4^2+..........-(n-1)^2+n^2=(-1)^{n+1}(\frac{n}{2})(n+1)$

For n=1:

$1^2=(-1)^2(\frac{1}{2})(2)$

$1=1$

Hence true for n=1.

Assume true for n=k:

$1-2^2+3^2-4^2+..........-(k-1)^2+k^2=(-1)^{k+1}(\frac{k}{2})(k+1)$

For n=k+1:

$1-2^2+3^2-4^2+..........-k^2+(k+1)^2=(-1)^{k+2}(\frac{k+1}{2})(k+2)$

$1-2^2+3^2-4^2+..........+(k-1)^2-k^2+(k+1)^2=(-1)^{k+2}(\frac{1}{2})(k^2+3k+2)$

$1-2^2+3^2-4^2+........+(k-1)^2-k^2+(k+1)^2=(-1)^{k+2}(\frac{1}{2})(k)(k+1)+(-1)^{k+2}(\frac{1}{2})(2k+2)$

$1-2^2+3^2-4^2+.........+(k-1)^2-k^2+(k+1)^2=-(-1)^{k+1}(\frac{1}{2})(k)(k+1)-(-1)^{k+1}(k+1)$

At this point it's clear that the LHS does not equal the right RHS. This implies that my initial expression for the sum is incorrect. However, it works for n=1 and n=2 so I think it's right.

Can anyone see what i'm doing wrong?

2. Originally Posted by Showcase_22
I tried this last night and i'm looking at it this mornng. I'm still stuck on a bit of it:

My method is like this:

$1-2^2+3^2-4^2+.........-(n-1)^2+n^2=(-1)^{n+1}(\frac{n}{2})(2+(n-1))$

$1-2^2+3^2-4^2+..........-(n-1)^2+n^2=(-1)^{n+1}(\frac{n}{2})(n+1)$

...
In my opinion you must use the factor $(-1)^{n+1}$ at the LHS too. Otherwise you must examine the equation for 2 different cases: n is even or n is odd.

3. How do you mean?

Like this:

$1-2^2+3^2-4^2+..........+(-1)^{n+1}(n-1)^2+(-1)^{n+1}n^2=(-1)^{n+1}(\frac{n}{2})(n+1)$

?

4. Exactly, since each even term has a negative sign in front of it which must be taken into account.

5. Hello, Showcase_22!

We have: . $1^2 - 2^3 + 3^2 - 4^2 + \cdots + (\text{-}1)^{n-1}n^2 \;=\;(\text{-}1)^{n-1}\frac{n(n+1)}{2}$

Verify $S(1)\!:\quad (\text{-}1)^01^2 \:=\:(\text{-}1)^0\frac{1(2)}{2}$ . . . True

Assume $S(k)\!:\quad 1^2 - 2^2 + 3^2 - 4^2 + \cdots + (\text{-}1)^{k-1}k^2 \;=\;(\text{-}1)^{k-1}\frac{k(k+1)}{2}$

Add $(\text{-}1)^k(k+1)^2$ to both sides.

. . $\underbrace{1^2-2^2+3^2 - 4^2 + \cdots + {\color{blue}(\text{-}1)^k(k+1)^2}}_{\text{Left side of }S(k+1)} \;=\;(\text{-}1)^{k-1}\frac{k(k+1)}{2} + {\color{blue}(\text{-}1)^k(k+1)^2}$

We must show that the right side is the right side of $S(k+1)$
. . which looks like this: . $(\text{-}1)^k\frac{(k+1)(k+2)}{2}$

The right side is: . $(\text{-}1)^{k-1}\frac{k(k+1)}{2} + (\text{-}1)^k(k+1)^2$

Factor: . $(\text{-}1)^{k-1}\cdot\frac{k+1}{2}\cdot\bigg[k + (\text{-}1)2(k+1)\bigg] \;=\;(-1)^{k-1}\cdot\frac{k+1}{2}\cdot[-k-2]$

Factor: . $(\text{-}1)^{k-1}\cdot\frac{k+1}{2}\cdot(\text{-}1)\cdot(k+2) \;=\;(\text{-}1)^k\cdot\frac{k+1}{2}\cdot(k+2)$

. . And we have: . $(\text{-}1)^k\frac{(k+1)(k+2)}{2}\quad\hdots\quad \text{ta-}DAA!
$

The inductive proof is complete.

6. Originally Posted by Showcase_22
How do you mean?

Like this:

$1-2^2+3^2-4^2+..........+(-1)^{n+1}(n-1)^2+(-1)^{n+1}n^2=(-1)^{n+1}(\frac{n}{2})(n+1)$

?
Quite. The exponent of the factor (-1)^{n+1} is exactly greater than n by 1. Therefore the LHS should be:

$1-2^2+3^2-4^2+..........+(-1)^{\bold{{\color{red}n}}}(n-1)^2+(-1)^{n+1}n^2=(-1)^{n+1}(\frac{n}{2})(n+1)$

7. ## Grouping terms

I would like to give a different way for the proof.
It is not by induction, its by direct computation.
This may be useful for other problems.

We want to prove that
$\sum\limits_{j=1}^{n}(-1)^{j+1}j^{2}=(-1)^{n}\sum\limits_{j=1}^{n}j=(-1)^{n}\frac{n(n+1)}{2}$ for any $n\in\mathbb{N}$.
I will only consider the case where $n\in\mathbb{N}$ is odd because the case where $n$ is even is very similar.
Since $n$ is odd, we may find $k\in\mathbb{N}$ such that $n=2k-1$ holds.
$\sum\limits_{j=1}^{2k-1}(-1)^{j+1}j^{2}=1+\sum\limits_{j=1}^{k-1}\big[(2j+1)^{2}-(2j)^{2}\big]$
.................... $=1+\sum\limits_{j=1}^{k-1}\big[(2j+1)^{2}-(2j)^{2}\big]$
.................... $=1+\sum\limits_{j=1}^{k-1}\big[4j+1\big]$
.................... $=k+4\sum\limits_{j=1}^{k-1}j$
.................... $=k+4\frac{(k-1)k}{2}$
.................... $=\frac{2k}{2}+\frac{(2k-2)2k}{2}$
.................... $=\frac{(2k-1)2k}{2}$
.................... $=\frac{n(n+1)}{2}.$
Hence the proof for this case is completed.
Let $n$ be even and pick $k\in\mathbb{N}$ to satisfy $n=2k$.
Then, we have
$\sum\limits_{j=1}^{2k}(-1)^{j+1}j^{2}=-\sum\limits_{j=1}^{k}\big[(2j)^{2}-(2j-1)^{2}\big]$.
The rest is similar.

8. Oh WOW!

I wasn't expecting quite so much help. I was trying it myself and I was still getting a bit stuck on it (my typo when I wrote $(-1)^{n+1}$ was really throwing me).

Anyway, thanks a bundle. I'm going to attempt this question and solve it like there's no tomorrow!!!

After trying and SUCESSFULLY solved the problem: I used a method similar to Soroban's except I did it a little differently:

$(-1)^k \frac{(k+1)(k+2)}{2}+(-1)^k(k+1)^2-(-1)^k (k+1)^2$

After a lot of working you end up with the required result (edit: I just noticed this is exactly what Soroban did. Rather than add it to both sides I added it to the same side).

I like your method bkarpuz. Direct computation is a method I haven't heard of before. It took me a few read-throughs but I finally got it. =D

9. Originally Posted by Showcase_22
Oh WOW!

I wasn't expecting quite so much help. I was trying it myself and I was still getting a bit stuck on it (my typo when I wrote $(-1)^{n+1}$ was really throwing me).

Anyway, thanks a bundle. I'm going to attempt this question and solve it like there's no tomorrow!!!

After trying and SUCESSFULLY solved the problem: I used a method similar to Soroban's except I did it a little differently:

$(-1)^k \frac{(k+1)(k+2)}{2}+(-1)^k(k+1)^2-(-1)^k (k+1)^2$

After a lot of working you end up with the required result.

I like your method bkarpuz. Direct computation is a method I haven't heard of before. It took me a few read-throughs but I finally got it. =D
Direct computation is not a method, its just calculating the desired identity directly by using well-known identities.
However, you may not always come across with a known one and you may need to provide your own (new) identities.