Solving Math Induction Problem from Apostol's Calculus vol1 pg35/36

[enian]

This problem is from Apostol's Caclulus book vol 1 page 35 or 36 #2

Show that
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4)
is true by mathematical in duction

I get to this step but have problem figuring out how to finish it off

-1^(n+1) * n^2 = -1^(n+1) * (n(n+1)/2)

I then do the plugging in bit and end up with

-1^(n+1) * (n(n+1)/2) + (-1^(n+2) * (n+1)^2) << the next term (n+1)

I have tried to work this out a few times but I can't seem to get a stable answer, any helps? Maybe I've forgotten some simple algebra tricks. The answer I think should look like

-1^(n+2) * ((n+1)(n+2)/2)

 

[matt grime]

Steady on, fella.

-1^(n+1) * n^2 = -1^(n+1) * (n(n+1)/2)

is clearly not a good thing to write. If you were just given that you can solve for n, so that isn't what you're supposed to write.

This is induction. So we're going to show that P(n) implies P(n+1), right? What is P(n)? What things are involved in it, and how can we relate P(n) to P(n-1)?

 

[enian]

If I was just given that I can solve for N? What do you mean?

That's the pattern that I pulled from the set of equations..
1 - 4 = -(1 + 2)
1 - 4 + 9 = 1 + 2 + 3
1 - 4 + 9 - 16 = -(1 + 2 + 3 + 4)
is true by mathematical induction

So I got this mess
1 - 4 + 9 - 16 + ... + -1^(n+1) * n^2 = -1^(n+1) * (n(n+1)/2)

I add the next term to the other side in an attempt to show that P(n+1) is ... something, valid? I have no idea.. I just know it's supposed to look like -1^(n+2)*((n+1)(n+2)/2), which i'll call F()

-1^(n+1) * (n(n+1)/2) + (-1^(n+2) * (n+1)^2) is where I'm stuck, trying to factor/expand/add/whatever I just end up with a mess that looks similar to F()

Mathematical Induction is so confusing. I don't know why I'm not getting it. I have read every stinking article I can find on this. Do I not just add the next term in line to both sides and solve so it looks like f(n+1) ??

 

[Dick]

Carefully take the difference between the (n+1)th form on the left side and the nth form on the left side and ditto on the right side. On the left you'll get a sign times (n+1)^2. On the right you will get the same sign times (n+2)(n+1)/2+(n+1)n/2. You were already halfway there before you got angry. If you can show they are the same then you win. That's stinking mathematical induction, and it wasn't that bad, now was it? The problem that you had -1^(n+1) * (n(n+1)/2) + (-1^(n+2) * (n+1)^2) and you should have had the difference -1^(n+1) * (n(n+1)/2) - (-1^(n+2) * (n+1)^2). Do you see why?

 

[matt grime]

INduction is simple. I want to prove that

1-4+9+...+(-1)^{n+1}n^2

equals something, which we'll call P(n). P(n) is also equal to

P(n-1)+ (-1)^{n+1}n^2

and by induction we have a formula for P(n-1), so use it amd rearrange.

 

[enian]

Dick:
I don't see why I would subtract the last term -(-1^n+2...) Because if you distribute the negative sign you're left with 1^n+2 which would always be positive, even if the number before it were positive, which contradicts the changing sign on the even/odd of the general equation. So no, I don't see why. :(

Maybe this is just too over my head.

Matt Grime: Why do you use P(n-1)? Simpler math? Since it will cancel out the -1^[n+1] in front of the terms? Or is there something I'm missing between using P(n-1) or P(n+1)

 

[matt grime]

Matt Grime: Why do you use P(n-1)?

Because that is what induction is. You have something to prove, something for each natural number, say. So to show it is true for some n, you write the statement for n in terms of (some of the) statements for 1,2,..,n-1 which you assume to be true. This allows you to deduce the result you want is true.

Then it only remains to show it is true for the first case.

That is induction. Every induction argument is based upon that premise. The first thing you have to think about in induction is 'OK, how can I relate the statement for the n'th thing to the statements for the previous ones...'

I may have misunderstood you: if your question was 'why did you use n-1 to show n, rather than n to show n+1' I hope you understand that there is no difference. OK, some times things are easier to manipulate if you do this, but here it is no difference. It is just an algebraic manipulation you have to perform.

 

[dextercioby]

Ok, let's show that P(n)-->P(n+1). Which is basically reduced to the formula

$$1-4+9-16+...+(-1)^{n+1}n^2 +(-1)^{n+2} (n+1)^2 =(-1)^{n+1}\frac{n(n+1)}{2}+(-1)^{n+2} (n+1)^2$$

So do the calculation of the RHS. I'll give you 2 hints:
*Factor the (n+1).
* at some point you'd have to factor a "-1".

 

[enian]

thanks dexter. That's the step where I got lost.
https://www.physicsforums.com/latex_images/14/1413195-0.png

The first thing I do is put the second term on the right hand side of the equation, over 2, to get a common denominator so I can add them together. I'm left with..

-1^[n+1]*n(n+1) - 1^[n+2]*2(n^2+2n+2), all over 2

This is the part I have trouble.
Can -1^[n+2] be multiplied by 2^1? So I would get -2^(n+2)?
I'm not sure if the [n+2] exponent can be combined with a 1 exponent, similar to trying to add 2x+2. Am I right with that thinking?

Also, is it possible to factor a -1^[n+1] from a equation such as...
-1^[n+1] + -1^[n+2] ? So that it would look like..
-1^[n+1](1 + (1/(1^n+1)))

The main thing that trips me up here is the exponents.

What do you use to make the nice equations (the graphical ones as opposd to my text ones) dexter?

 

[matt grime]

My 2 penn'orth: get your brackets in the right place. If you keep making mistakes like that then you'll never get to the maths.

tCan -1^[n+2] be multiplied by 2^1?

2^1 is just 2, of course you can multiply a number, (-1)^{n+2} (which is 1 if n is even and -1 if n is odd, and notice the brackets!), by 2.

So I would get -2^(n+2)?

of course not. Firstly, this number is now always negative, and goes like -2,-4,-8,-16,..

I'm not sure if the [n+2] exponent can be combined with a 1 exponent,

you can only add exponents if the things that are being exponentiated are equal. -1 does not equal 2.

-1^[n+1] + -1^[n+2] ? So that it would look like..
-1^[n+1](1 + (1/(1^n+1)))

I'm going to say it again: bracket things properly. Forget it's -1, you're asking to simplify x^{n+1} + x^{n+2}, which is just x^{n+1}(1+x), though I don't think that is quite what you should be asking to do.

 

[matt grime]

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

That equals

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

right?

 

[enian]

The main thing that I missed on this problem was the fact I could break down an exponent into two separate parts via x^(a+b)=x^a*x^b
Thanks for the assistance.