Maths question

I think for part (iii) you are supposed to infer that the five-term approximation tends to 1 + 1 + 1/2 + 1/6 + 1/24 for very large values of n. There may be some words missing in the mark scheme.

In part (iv) the thing about e is that if you have come across the numerical value of e before, you may recognise something rather close to it it in the spreadsheet for very large values on n.

As for filling out the tables, it's easy enough in a spreadsheet as you can simply copy the relevant spreadsheet formula from one cell to another. On a calculator like the Casio FX-991EX you can save a bit of time by using Table Mode, or in normal mode by typing in expressions for the third, fourth and fifth terms or the series separate by "ALPHA :" symbols then CALC'ing for different numerical values.

Thanks very much for your reply
Sorry I have just attached here the correct version:
And how did you infer for part iii) that the five-term approximation tends to 1 + 1 + 1/2 + 1/6 + 1/24 for very large values of n?

And how did you infer for part iii) that the five-term approximation tends to 1 + 1 + 1/2 + 1/6 + 1/24 for very large values of n?

For a fraction like $\dfrac{(n-1)(n-2)}{6n^2}$ it is easy to see that if you expand the numerator and take the highest order term, which is $n^2$ here, then divide each term of the fraction by it, then you will be left with $\dfrac{1-\frac{3}{n}+\frac{2}{n^2}}{6}$. When $n \rightarrow \infty$, each little fraction with an $n$ in the denominator will tend to 0, hence the limit is $\frac{1}{6}$

For example ((n - 1)(n - 2)) / 6n^2 = (n^2 - 3n + 2) / 6n^2. For very large n, the "3n" and "2" terms will become vanishingly small compared to the x^2 terms. So (n^2 - 3n + 2) / 6n^2 --> n^2 / 6n^2 --> 1 / 6.
Similar reasoning for the other terms.

Thanks very much -the calculations side of this question makes sense now.
I am a bit confused with the wording of the mark scheme:

As n get bigger, all the “ratios” of terms involving n in the approximation get closer and closer to 1, so in the limit the sequence of the Professor’s calculations seem likely to tend to 2 17/24.

Doesn’t it get closer and closer to 0 b/c i.e. ((n-1)(n-2))/6 = (1- (3/n)-(2/n^2))/6

-3/n and 2/n^2 get closer and closer to 0 as n-->infinity so 1-0/6 --> 1/6

More like the limit of the sequence is 2 17/24 or the sequence tends to 2 17/24 not “in the limit the sequence of the Professor’s calculations seem likely to tend to 2 17/24.”

Could you please let me know if I have the wrong understanding for the wording of the mark scheme answer?

I also found the wording in the mark scheme a little odd. Here's a stab at what the words might mean: "as n gets bigger, the ratio of the value of any particular term in the series for n = (k+1) to the value for n = k approaches 1, that is to say the value of the term changes more slowly as n gets larger and in fact tends towards a fixed value for very large n".

For a fraction like $\dfrac{(n-1)(n-2)}{6n^2}$ it is easy to see that if you expand the numerator and take the highest order term, which is $n^2$ here, then divide each term of the fraction by it, then you will be left with $\dfrac{1-\frac{3}{n}+\frac{2}{n^2}}{6}$. When $n \rightarrow \infty$, each little fraction with an $n$ in the denominator will tend to 0, hence the limit is $\frac{1}{6}$

If you look at the $\dfrac{(n-1)(n-2)}{6n^2}$ term, for example, you can write this as:

$\left(\dfrac{n-1}{n}\right) \left(\dfrac{n-2}{n}\right) \dfrac{1}{6}$

The two items in brackets are the ratios (n-1)/n and (n-2)/n, both of which tend to 1 as n tends to infinity.

(Your own argument is also valid (but different), but you've had to multiply out for it, and obviously that will get painful when you have enough terms multiplied together).

Edit: FWIW, I'd say this is a somewhat sloppily worded question and solution - it's unfortunate that the person writing it doesn't seem quite able to decide whether readers are supposed to understand limits or not, which I think makes things more confusing. Where's the Q from?

It's from Mei question papers BTW this is Core 1

Are you self-teaching from these resources?

Well kind of, my teachers are not that great.

From what I've seen with your questions, these resources aren't that great either - they're not as clear, precise and explained as they should be.