# How do you do a proof on this?

1. Jul 15, 2011

### gregorious

1. The problem statement, all variables and given/known data

1+13+37+73...+[(n/2)(2+(n-1)12)-n+1]= [(n/2)(2+(n-1)12)-n+1]

2. Relevant equations
The numbers are actual data that I found and the equation on the right side is the general formula I found. Am I setting up the proof wrong by repeating the formula on the left side.

3. The attempt at a solution
I tried to do an induction proof. But because the Formula on the left side is a sum of everything before it I don't think it can be proofed because the right side is the same equation. Is there any other way to test the validity of the statement?

2. Jul 15, 2011

### Staff: Mentor

Your generating function $\frac {n} {2} ( 2+(n-1)12 ) -n+1$
looks like it can be simplified. The numerator and denominator there are both multiples of 2.
Then remove the brackets, and you see an n minus n which cancel. So it can be tidied up to something less fearsome.

I don't know whether this makes things easier, or whether there is a reason for the expression being written as it is.

3. Jul 16, 2011

### Ray Vickson

The term t(n) = (n/2)(2 + 12(n-1)) - n+1 is the nth term of your sequence (assuming the second differences of terms continue to be 12 for higher n). It is not the SUM of the first n terms, as you apparently seem to be saying (or are you? it is not clear). You can simplify t(n) a lot, then apply known (or tabulated) sums to get the final result. Alternatively, you can set sum[t(k),k=1..n] = a + b*n + c*n^2 + d*n^3, then determine the values of the constants a, b, c and d by looking at explicit numerical values of the sum for n = 1, 2, 3 and 4, and determining what a, b,c and d you need to get those values.

RGV

4. Jul 16, 2011

### gregorious

I'll try to explain better here. Basically I was given the first three terms 1, 13, 37. To start with. Then I had to find a formula that would give those values.
I started with the summing of an arithmetic sequence formula.
Sn = (n/2)(2u1 + (n − 1)d ) which if u try by itselt u'll see it doesnt work. So I modified it with the nth term of an arithmetic sequence un = u1 + ( n − 1)d to get the final formula I gave earlier that actually works with all the values. Now im asked to test the validity of the final formula I came up with. However I'm not sure if thats actually possible or if theres a method other than induction which I can use. In my attempt to prove it i Set up like this
1+13+37+73...+[(n/2)(2+(n-1)12)-n+1]= [(n/2)(2+(n-1)12)-n+1]. Using the one formula on both sides of the equation where usually on the right side u would have a sum formula (which i have) and on the left would be a (U sub n) formula which in my case is the same as the sum. I tried to solve it anyway (yes i did simplify and all that jazz) but the numbers wouldn't come out. Now im asking if anyone else can do a proof because I can't and would u please post it here. Or is it impossible like im thinking because both sides have the same formula? If so is there any other way to prove the validity of the formula without induction perhaps.

5. Jul 16, 2011

### Char. Limit

Something I like to look at is differences. For example, let's take the first four terms:

a1=1, a2=13, a3=37, a4=73

All right, now define bn=an+1-an.From this, we can take three terms of bn:

b1=12, b2=24, b3=36

And defining cn=bn+1-bn, we see that cn is constant:

c1=12, c2=12.

What's that tell me? Well, cn is constant, bn is linear, so an must be points on a quadratic in n. From there it's just figuring out which quadratic it is.

6. Jul 16, 2011

### gregorious

@Char. Limit
Thats about the first thing I did... Hence the 12 you found being at the d spot in my equation. What Im asking is if its actually possible to do a proof with the same equation in both spots. I.e: Solve for n = 1 (check). Do k+1 (numbers come out wrong for meE). Did i make a mistake somewhere >-<

7. Jul 16, 2011

### Ray Vickson

You are doing things the hard way. A direct approach is quicker and easier. Let t(n) = nth term, let d1(n) = t(n+1) - t(n) (first difference) and d2(n) = d1(n+1) - d2(n) (second difference). We see that the first few second differences d2(n) are all 12. Assuming that holds for all n, we have: d1(n+1) = 12 + d(n) = 2*12 + d(n-1) = ..., so d1(n) = a + 12n for some constant a. Since d1(1) = 13 - 1 = 12, we have a = 0. Now we know that t(n+1) - t(n) = 12n, so t(2) = t(1) + 12, t(3) = t(2) + 12*2 = t(1) + 12*[1+2],.... t(n) = t(1) + 12*(1 + 2 + ... + (n-1)). This last sum is standard, and can be found in many books or tables, and can even be found on-line. However, there is an easy way to get it: let S = 1 + 2 + ... + (n-1). We have 2S = [1 + (n-1)] + [2 + (n-2)] + ... + [(n-1) + 1] (just listing the terms in one of the S in ascending order and the other in descending order). We have (n-1) terms, each of which equals n, so 2S = n(n-1), hence S = n(n-1)/2. So, we have t(n) = t(1) + 6*n*(n-1). We have t(1) = 1, so the general term is t(n) = 1 + 6*n^2 - 6*n.

RGV

8. Jul 16, 2011

### vela

Staff Emeritus
First off, what exactly are you trying to prove? From what you've written so far, you've been given the first few terms of a sequence and asked to find a general formula for that sequence, which you found.

If you're trying to show the formula works, just plug in n=1, n=2, n=3, etc. and show it gives the correct results.

It's not clear why you're adding the terms in the sequence. What you wrote:
is obviously not true because if you subtract [(n/2)(2+(n-1)12)-n+1] from both sides, you get 1+13+... = 0.