Mega A Level Maths Thread MK II

OP

Original post by joostan

Well, what's d, and a? Subbing those values into the derivation from the start (or the end) can yield the answer - alternatively I could prbably dream up a proof by induction :smile:

No, it's fine...

a = 1
d = 4

n/2 (1 + 4(n-1))

(n + 4n^2 - 4n)/2
= 4n^2 - 3n/2

So I'm confused

Reply 843

Original post by L'Evil Fish

No, it's fine...

a = 1
d = 4

n/2 (1 + 4(n-1))

(n + 4n^2 - 4n)/2
= 4n^2 - 3n/2

So I'm confused

Check the bold - you'll kick yourself when you spot it :tongue:

Reply 844

OP

Original post by joostan

Check the bold - you'll kick yourself when you spot it :tongue:

(n-1) should be substituted with the other value they have? (4n-7)??

Reply 845

Original post by L'Evil Fish

(n-1) should be substituted with the other value they have? (4n-7)??

You're over complicating this :lol:

Spoiler

Reply 846

OP

Original post by joostan

You're over complicating this :lol:

Spoiler

Sn = n/2 (2a + (n-1)d)
= n/2 ((2)(1) + (n-1)4)

Don't get what I've done wrong

Reply 847

Original post by L'Evil Fish

Sn = n/2 (2a + (n-1)d)
= n/2 ((2)(1) + (n-1)4)

Don't get what I've done wrong

Precisely.
What's

2 \times 1

? and check your working :smile:

Reply 848

OP

Original post by joostan

Precisely.
What's

2 \times 1

? and check your working :smile:

That'll happen in the exam too...

n/2 (2 + 4n-4)
n/2 (-2+4n)
(-2n+4n^2)/2
2n^2-2n
2n(n-1) or even n(2n-1) as they want

(edited 10 years ago)

Reply 849

Fuzzy12345

10

Original post by L'Evil Fish

...

They're not equivalent, I suspect you've made an error somewhere.

Don't really understand what the question means and what constitutes as deriving from first principles, but alternatively you could always split the sum up to:

\displaystyle\sum_{r=1}^{n} 4r - 3 = \displaystyle\sum_{r=1}^{n} 4r - \displaystyle\sum_{r=1}^{n} 3

and substitute

4r = \frac{1}{2} ([2r+1]^{2} - [2r-1]^{2} )

which all telescopes to

\frac{1}{2}((2n+1)^{2} - 1) - 3n

and simplify

(edited 10 years ago)

Reply 850

Original post by L'Evil Fish

That'll happen in the exam too...

n/2 (2 + 4n-4)
n/2 (-2+4n)
(-2n+4n^2)/2
2n^2-2n
2n(n-1) or even n(2n-1) as they want

I'm sorry to say that the two forms you've given me are not equivalent. Check the bold.
The inductive proof was so much tidier :tongue:

Reply 851

username937760

16

Original post by Boy_wonder_95

Ouch how many marks did that cost you? I dropped 1 or 2 marks at the end as I forgot to times the probability by 2 :blushing:

, but there was a lot of tricky parts!

This question buggered me:

S1 question.png

I got the right answers using trial and error but what would be a more neater way of solving it?

As said, it cost me 5 UMS - I don't know exactly how many raw marks that is though, and I'm fairly sure I got the rest of the paper right.

Original post by L'Evil Fish

I.don't normally do this. But help please :frown:

You're not allowed to just quote the summation formula, so there are a few things you could do. Firstly, you could prove the summation formula

S_n = \frac{1}{2}n(2a+(n-1)d)

(I think that's the formula) but in my opinion a better way would be to simply prove the given summation using induction.

(You could also prove the result for the sum from one to n of r, and then use this and series methods to derive the result. The problem is that I'm not familiar with the WJEC syllabus so I don't know what method is going to be expected.)

Reply 852

Original post by Fuzzy12345

Don't really understand what the question means and what constitutes as deriving from first principles, but you could always split the sum up to:

\displaystyle\sum_{r=1}^{n} 4r - 3 = \displaystyle\sum_{r=1}^{n} 4r - \displaystyle\sum_{r=1}^{n} 3

and substitute

4r = \frac{1}{2} ([2r+1]^{2} - [2r-1]^{2} )

which all telescopes to

\frac{1}{2}((2n-1)^{2} - 1) - 3n

and simplify

Is this the same problem L'Evil Fish is working on? :eek:

Check the last few posts.
There's a proof by induction, but I think that this is satisfactory. Also I'm not convinced that what you've done is from first principles :s-smilie:

EDIT: I realise now that you were trying to help L'Evil Fish :smile:

(edited 10 years ago)

Reply 853

Original post by DJMayes

You're not allowed to just quote the summation formula, so there are a few things you could do. Firstly, you could prove the summation formula

S_n = \frac{1}{2}n(2a+(n-1)d)

(I think that's the formula) but in my opinion a better way would be to simply prove the given summation using induction.
(You could also prove the result for the sum from one to n of r, and then use this and series methods to derive the result. The problem is that I'm not familiar with the WJEC syllabus so I don't know what method is going to be expected.)

The formula is right - induction so much tidier but I'm pretty sure it's on the FP1 spec :tongue:

Reply 854

justinawe

19

Original post by Boy_wonder_95

Just did the S1 paper you sat, it was quite greasy...

Which paper was that? :lol:

Reply 855

Fuzzy12345

10

Original post by joostan

Is this the same problem L'Evil Fish is working on? :eek:

Check the last few posts.
There's a proof by induction, but I think that this is satisfactory. Also I'm not convinced that what you've done is from first principles :s-smilie:

EDIT: I realise now that you were trying to help L'Evil Fish :smile:

Yeah...I was trying to help the OP. :tongue:

Also really? :s-smilie:

You've assumed the summation formula in your solution and the question explicitly asks for a solution from first principles and I'd argue that the method of differences is as elementary as you can get - it's just toying around with algebra...

Reply 856

username937760

16

Original post by justinawe

Which paper was that? :lol:

Jan 2012, if he's correct that it's the one I sat.

Reply 857