Find the sum of a sequence, then prove with induction

darksteel88
Messages
4
Reaction score
0

Homework Statement


Given the sequence 1+5+9+...+(4n+1)

Find the sum of the sequence and then prove that it holds for all natural numbers using induction.


Homework Equations


Just induction, that we solve it using induction and, according to the teacher, by trial and error.

Induction states that we solve the base case (that it works at n=1).
Then we assume that it works for k which is an element of the natural numbers
Then we prove that if k = k+1 that it works and by induction, it works for all values of k

The Attempt at a Solution


I used the equation for the sum of a arithmetic sequence to figure out what the sum of that equation is. I went to http://en.wikipedia.org/wiki/Arithmetic_progression

Using the arithmetic sum equation I ended up with

S = n(2n - 1)

but clearly at n=1 we get

4n+1 = n(2n-1)
5 = 1

The teacher states that the equation is entirely correct, that is to say, there is some sort of sum that will work. I just used the math for it to prove that it won't work so I am completely lost.
 
Physics news on Phys.org
After much deliberation with another friend of mine, we came to this conclusion:

If we denote the 1st term by n=0, the 2nd term by n=1 and so on we can use

4n+1 = (n+1)(2n+1)

4(0)+1 = 1 (0+1)(2(0)+1) = 1

and so on, we can see it works

But he explicitly defined in class that in our class, 0 is not an element of the natural numbers.
It also works if we do other things like start the sequence at 5 and get a different sum or change it to 4n-3 instead of 4n+1

Thoughts?
 
You could prove 0 as a special case, and then proceed to do induction on the Naturals (which you did) if you’re afraid your teacher will gripe about 0 not being a Natural.

But in general you can do induction starting at any integer.
 
Cool, it works for n = 0 too, but that doesn't stop you from ignoring that and starting your induction from n = 1. Usually, you don't get marked down for proving more than you need to.
 
Well, my initial concern is that 4n+1 doesn't represent the nth term in the sequence as it should. It only represents it if we start from the notation that the first term is the 0th term in the sequence. It should be 4n-3 instead.

And he's made it clear we don't include 0 in the natural numbers because we've done questions that rely on that in class. Of course, I may have to. Thankfully I've got some time before I need the answer so I can see what some TAs say about it but still.

His reply just now was:

"Your formula may work for n=0 (which is not a natural number in our course). However, you can deal with n=1,2,3,... only (you don't have to cover the case n=0).
Also, we didn't cover the topic of arithmetic sequences in class. The question can be solved with induction only."

So I suppose as long as the induction works, then it's a sound question (despite other mathematics such as arithmetic sequences disproving the results. If I do the arithmetic series sum with this sequence starting at 0, I believe it gives a term that I can use induction on and find that it works out.

Thanks for the help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top