sum of n terms

tony123 · 2007-12-11 07:29:56

Find the sum of n terms of the series

1 + 8 + 19 + 34 + 53 + 76 + ------

mathsyperson · 2007-12-11 07:55:14

Identity · 2007-12-11 13:58:10

Let

denote the xth term

As the differences become constant after the 2nd time, the expression is a quadratic.

After exhaustive simultaneous equations we come up with

So now we wanna sum that:

Which is what mathsyperson got.

Last edited by Identity (2007-12-11 21:46:41)

mathsyperson · 2007-12-11 14:18:13

Yeah, I probably should have said all that stuff instead of being lazy.
I've got a different way of finding the nth term of that sequence though.

After determining that the 2nd difference is always 4, that means that the nth term of the sequence must contain 2n², because a constant ith difference k of a sequence means that the nth term contains (k/i!)n^i. (This may be an inefficient method, I discovered this pattern when I was about 10 and haven't grown out of using it to solve stuff like this. It also only works if you're given a chain of consecutive terms - if there are unknown gaps then you start to get problems).

So next we take away 2n² from the original sequence to get a new one. In this case we have:
-1, 0, 1, 2, 3, ...

If you're being robotic about it, you could take differences, notice that the 1st difference is always 1 and then use the above pattern to deduce that the nth term contains "n", but you really could have just got that by recognition anyway.

And then similarly, taking away n from that sequence leaves you with -2, -2, -2, -2, ... and so the constant term is obvious.

Ricky · 2007-12-11 14:22:27

To avoid simultaneous equations and a more general way of finding the equation of a series is using a method called guess and subtract.

Looking at the numbers in the series and guess that it's probably a polynomial (based on how fast they increase). So I'm going to start with a quadratic and move on from there. My first guess is that it's n^2. So we subtract n^2 from your series to see what that gives us:

Your series 1 8 19 34 53 76
n^2 1 4 9 16 25 36
Difference: 0 4 10 18 28 40

Now I take note that each time the difference is almost 1/2 of what the number in the series was. So I double my n^2

Your series 1 8 19 34 53 76
2n^2 2 8 18 32 50 72
Difference: -1 0 1 2 3 4

It should be clear what I'm missing is an n term, so I'm going to subtract n from this:

Your series 1 8 19 34 53 76
2n^2 + n 3 10 21 36 55 78
Difference: -2 -2 -2 -2 -2 -2

Now all that is needed is to tack a -2 on the end, and we have two sequences that are equal.

2n^2 + n - 2

If I had tried n^2 and found that it wasn't increasing fast enough, I would have tried n^3, and then n^4. Like wise, a similar method known as guess and divide can be used, but only when you believe your series has to do with an exponential function such as 2^x.

Math Is Fun Forum

#1 2007-12-11 07:29:56

sum of n terms

#2 2007-12-11 07:55:14

Re: sum of n terms

#3 2007-12-11 13:58:10

Re: sum of n terms

#4 2007-12-11 14:18:13

Re: sum of n terms

#5 2007-12-11 14:22:27

Re: sum of n terms

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