Finding the nth term of 2,1,3,0,4,-1,5,-2,6,-3,...

I

inneedofmathshelp

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Hi,
Given this sequence of numbers how does one find an expression for the nth term?

It's a question set for my daughter's lockdown work and I'm stumped! We see the differences between each term as 1, 2, 3, 4,etc (or should it be -1, 2,-3, 4 etc?) but we've no idea how to use that to generate an nth term expression.

We'd be so grateful for any help folk here might be able to provide.
 
inneedofmathshelp said:
Hi,
Given this sequence of numbers how does one find an expression for the nth term?

It's a question set for my daughter's lockdown work and I'm stumped! We see the differences between each term as 1, 2, 3, 4,etc (or should it be -1, 2,-3, 4 etc?) but we've no idea how to use that to generate an nth term expression.

We'd be so grateful for any help folk here might be able to provide.
Click to expand...
Pleas let us know:

Your daughter's grade-level and

the topic of the math-class being discussed now (or just prior to this problem) and

it will be better if your daughter could communicate with us directly.
 
Hi,

Dau is 13 and the topic being set is simply a challenge problem set as part of home study during lockdown and not part of any particular unit of study.

HTH
 
inneedofmathshelp said:
Given this sequence of numbers how does one find an expression for the nth term?

It's a question set for my daughter's lockdown work and I'm stumped! We see the differences between each term as 1, 2, 3, 4,etc (or should it be -1, 2,-3, 4 etc?) but we've no idea how to use that to generate an nth term expression.

We'd be so grateful for any help folk here might be able to provide.
Click to expand...
You have done well so far.

One hint that may help is that the answer may come in two parts, one for even n and one for odd n. Don't expect to have a single formula (at least not initially).

Of course, make sure you are doing as we try to do, guiding without doing too much for her.
 
Thanks, yes, she had worked out the sequence for even and for odd and the need for an expression for each alternate number in the sequence but has only seen other examples where the difference between each value works on each consecutive value in the sequence.

What might such an expression look like? I'm grateful for the pointers and seeing an alternative expression would be most helpful to aid her in applying something similar to the given sequence - i.e. no answer needed (I hope :) )

She had also wondered if it might be quadratic and the expression needed to look at the differences of differences (if I've explained that properly).
 
What did she work out? I want to start with that.

It is not quadratic. The sequence of odd terms is 2, 3, 4, 5, 6, and the sequence of even terms is 1, 0, -1, -2, -3. Each is linear.

There is no need to combine these two formulas into one!! I'd like to see the exact wording of the problem as given, to make sure I know what is required. But in general, piecewise function definitions are perfectly legal.
 
Hi,

Many thanks for taking the trouble to reply.

The original problem posed: "Given the sequence of 2,1,3,0,4,-1,5 ... (a) find the product of the 20th and 21st term (b) can you work out the nth term?"

She's answered (a) as -8 * 12 -> -96

For (b) she's seen the linear alternate sequence and is quite familiar with problems i.e. 3n+1 (4,7,10,13,16 ...), or -3n (-3,-6,-9,-12 ...) but this one seems to be of a different order with two linear sequences alternating ...?
 
inneedofmathshelp said:
The original problem posed: "Given the sequence of 2,1,3,0,4,-1,5 ... (a) find the product of the 20th and 21st term (b) can you work out the nth term?"

She's answered (a) as -8 * 12 -> -96

For (b) she's seen the linear alternate sequence and is quite familiar with problems i.e. 3n+1 (4,7,10,13,16 ...), or -3n (-3,-6,-9,-12 ...) but this one seems to be of a different order with two linear sequences alternating ...?
Click to expand...

Let's get into details.

The even terms are 1, 0, -1, -2, ..., so the kth even term is 2 - k. The 20th term is the 10th event term: 2-10 = -8.

The odd terms are 2, 3, 4, 5, ..., so the kth odd term is 1 + k. The 21st term is the 11th odd term: 1+11 = 12.

So he answer for (a) is correct.

The kth even term is the nth = 2k'th term, so k = n/2, and the nth term is 2 - n/2 = (4 - n)/2.

The kth odd term is the nth = 2k-1'th term, so k = (n+1)/2, and the nth term is 1 + (n+1)/2 = (n + 3)/2.

[MATH]a_n=\left\{\begin{matrix}\frac{4-n}{2} & n\text{ even}\\ \frac{n+3}{2} & n\text{ odd} \end{matrix}\right.[/MATH]​

That's the formula, as far as I'm concerned. You could combine those into one formula, with a little effort, but I don't think it's worth it. Such combined expressions are generally harder to use. In particular, the word "formula" isn't even used in the problem; you just have to have a way to work it out. (I do wish it were clearer how far you need to go, since good students are likely to try too hard, and feel unsuccessful!)

But just for fun, the average of the two formulas is [(n + 3)/2 + (4 - n)/2]/2 = 7/4, and the terms are alternately increased and decreased from this by [(n + 3)/2 - (4 - n)/2]/2 = (2n - 1)/4, so the combined formula would be 7/4 - (-1)^n*(2n - 1)/4.

Checking this, for n=1 we get 7/4 + 1/4 = 2; for n=2, we get 7/4 - 3/4 = 1.
 
Dr.Peterson said:
Let's get into details.

The even terms are 1, 0, -1, -2, ..., so the kth even term is 2 - k. The 20th term is the 10th event term: 2-10 = -8.

The odd terms are 2, 3, 4, 5, ..., so the kth odd term is 1 + k. The 21st term is the 11th odd term: 1+11 = 12.

So he answer for (a) is correct.

The kth even term is the nth = 2k'th term, so k = n/2, and the nth term is 2 - n/2 = (4 - n)/2.

The kth odd term is the nth = 2k-1'th term, so k = (n+1)/2, and the nth term is 1 + (n+1)/2 = (n + 3)/2.

[MATH]a_n=\left\{\begin{matrix}\frac{4-n}{2} & n\text{ even}\\ \frac{n+3}{2} & n\text{ odd} \end{matrix}\right.[/MATH]​

That's the formula, as far as I'm concerned. You could combine those into one formula, with a little effort, but I don't think it's worth it. Such combined expressions are generally harder to use. In particular, the word "formula" isn't even used in the problem; you just have to have a way to work it out. (I do wish it were clearer how far you need to go, since good students are likely to try too hard, and feel unsuccessful!)

But just for fun, the average of the two formulas is [(n + 3)/2 + (4 - n)/2]/2 = 7/4, and the terms are alternately increased and decreased from this by [(n + 3)/2 - (4 - n)/2]/2 = (2n - 1)/4, so the combined formula would be 7/4 - (-1)^n*(2n - 1)/4.

Checking this, for n=1 we get 7/4 + 1/4 = 2; for n=2, we get 7/4 - 3/4 = 1.
Click to expand...
Thank you that's really helpful to see an expression combining the two terms into one and the working out - thank you for taking so much time and trouble, we're really grateful.
 
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