Further Mathematics: Post your doubts here!

[mrpaudel]

Re: The Further Mathematics Thread. Post your doubts here!

I already appeared all my xams..and completed my A levels...but still...Since, i have interest in maths..i want to study Further mathematics....but...appearing xams later is not sure.....i wanna study just for fun....!! so some1 plz give me the link .....or any pdf file....like a book..so that i can study them...!!

 

[Nibz]

Re: The Further Mathematics Thread. Post your doubts here!

This http://patrickjmt.com/ & this http://mathworld.wolfram.com/ may help you somehow. Other than these, use a Cambridge Endorsed book of Further Mathematics and practice past exam papers.

 

[abcde]

Use 'Proof by Induction' to verify the statement: "For all integral values of n, 2^(n+2) + 3^(2n+1) is exactly divisible by 7."

 

[OakMoon!]

 

[abcde]

Prove by direct deduction and induction that Σ (lower limit : r=1, upper limit : n) ap^r = [a(1-p^n)]/(1-p) for positive integral values of n.

Isn't there an error in the question? Since the lower limit is r=1, the first term of the series would be 'ap'. So we should be asked to prove that Σ ap^r = ap (1-p^n)/(1-p) instead. No?

Secondly,

Prove by induction that 2^n is greater than 2n for all integral values of n greater than 2.

 

[OakMoon!]

Prove by direct deduction and induction that Σ (lower limit : r=1, upper limit : n) ap^r = [a(1-p^n)]/(1-p) for positive integral values of n.

Isn't there an error in the question? Since the lower limit is r=1, the first term of the series would be 'ap'. So we should be asked to prove that Σ ap^r = ap (1-p^n)/(1-p) instead. No?

Secondly,

Prove by induction that 2^n is greater than 2n for all integral values of n greater than 2.

Agreed. It is a geometric progresstion and in the formula you have to multiply with the first term. Hence, you are right.

 

[OakMoon!]

@abcde: here is the second question:

 

[abcde]

Thank you!
When you write, " 2^k - 2 > 0 for every value of k greater than 0", I think it should be "for every value of k greater than 1(for it to be correct) or 2 (as per the question's requirement)".

 

[OakMoon!]

Actually it was 2 that I intended to write. I noticed it when I uploaded the question, but it wasn't worth correcting. You understood it anyways, and I knew that you will.

 

[abcde]

Prove the following by induction:

 

[OakMoon!]

Can you please check if you wrote the correct identity. This one doesn't even stands true for n=1.

 

[abcde]

Sorry for that. Here's the correct one:

 

[OakMoon!]

 

[ahzahra]

I really need help with june 2002 q10 please its urgent http://www.xtremepapers.com/CIE/Int.../9231 - Further Mathematics/9231_s02_qp_1.pdf

 

[ahsanxr]

alzahra, what's giving you a problem on it? It's a pretty simple one if you understand vector spaces. Or do you just want a solution? I can help with either.

 

[abcde]

AoA!
Prove by induction that Σ r^2 = 1/6 n(n+1)(2n+1).
The proof is fairly simple. I need help with the next part:

Find the sum of the squares of the first n positive odd integers.

Secondly, how can we solve this?

Verify that 5^5 ≡ 1 (Mod 11). Hence find the remainder obtained on dividing 5^1998 by 11.

 

[OakMoon!]

Subtract the sum of positive even integers from the sum of squares of all integers in such a way: Σr²-Σ(2r)² and find the answer.

Didn't get the second question. Mod 11 as in |11|?

 

[abcde]

Subtract the sum of positive even integers from the sum of squares of all integers in such a way: Σr²-Σ(2r)² and find the answer.

Didn't get the second question. Mod 11 as in |11|?

Σ(2r)² is larger than Σr² so the answer would come out as -ive, don't you think? :S
The way I solved it was to substitute (2n -1) in place of n in this formula: Σ r^2 = 1/6 n(n+1)(2n+1).
That way I get this answer: n/3 (2n - 1)(4n - 1) whereas the correct answer is n/3 (4n^2 - 1).

No. Mod 11 as in the check digit. I don't know how it can be solved.

 

[OakMoon!]

You'll need to just use different limits. To find the n number of ODD integers you'll use 2n integers because n number of them will be even.
So use 2n for the sum of all integers and then subtract sum to n of positive integers that will be represented by 2r.

Here, I have solved it for you.



Hope this helped.

 

[OakMoon!]

Mind the writing! Did it really quickly