How do you do a proof on this?

[gregorious]

Homework Statement

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

Homework 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.

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?

 

[NascentOxygen]

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.

 

[Ray Vickson]

Homework Statement

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

Homework 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.

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?

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

 

[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 doesn't 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 I am asked to test the validity of the final formula I came up with. However I'm not sure if that's actually possible or if there's 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 I am asking if anyone else can do a proof because I can't and would u please post it here. Or is it impossible like I am 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.

 

[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 b_n=a_n+1-a_n.From this, we can take three terms of b_n:

b₁=12, b₂=24, b₃=36

And defining c_n=b_n+1-b_n, we see that c_n is constant:

c₁=12, c₂=12.

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

 

[gregorious]

@Char. Limit
Thats about the first thing I did... Hence the 12 you found being at the d spot in my equation. What I am 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 >-<

 

[Ray Vickson]

@Char. Limit
Thats about the first thing I did... Hence the 12 you found being at the d spot in my equation. What I am 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 >-<

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

 

[vela]

Thats about the first thing I did... Hence the 12 you found being at the d spot in my equation. What I am 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 >-<

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:

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

is obviously not true because if you subtract [(n/2)(2+(n-1)12)-n+1] from both sides, you get 1+13+... = 0.