by writing
n
∑ (r+1)2^(r-1) in terms of 2^r form a conjecture for the sum of this equation.
r=1
BTW: how do i partial fractions this?
1 ÷ {(2r -1)(2r+1)(2r+3)} as i need to prove this = 1/4[ 1/{(2r-1)(2r+1)} - 1/{(2r+1)(2r+3)} ]
tip:
1 way to 'play cheat' is to work backwards... So you sum up the partial fractions on the right hand side, then observe how it was done
Or you try expanding the right hand side partial fractions, then expand left hand side, then observe if there's any pattern linking them
Hi,
Substitute n with different values and use the hint in the question, i.e in terms of 2^r:
n = 1, S_1 = 2
n = 2, S_2 = 8
n = 3, S_3 = 24
n = 4, S_4 = 64
n = 5, S_5 = 160
Since we need the sum to be expressed in terms of 2^r, we try to express the above values in the form a (2^n) + b, where a and b are constants.
S_1 = 2 = 1 (2^1)
S_2 = 8 = 2 (2^2)
S_3 = 24 = 3 (2^3)
S_4 = 64 = 4 (2^4)
S_5 = 160 = 5 (2^5)
So S_n = ? (2^?)
I guess you can then proceed from this point ya?
P.S. The question does not seem to be very clear.
Thanks!
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
Substitute n with different values and use the hint in the question, i.e in terms of 2^r:
n = 1, S_1 = 2
n = 2, S_2 = 8
n = 3, S_3 = 24
n = 4, S_4 = 64
n = 5, S_5 = 160
Since we need the sum to be expressed in terms of 2^r, we try to express the above values in the form a (2^n) + b, where a and b are constants.
S_1 = 2 = 1 (2^1)
S_2 = 8 = 2 (2^2)
S_3 = 24 = 3 (2^3)
S_4 = 64 = 4 (2^4)
S_5 = 160 = 5 (2^5)
So S_n = ? (2^?)
I guess you can then proceed from this point ya?
P.S. The question does not seem to be very clear.
Thanks!
Cheers,
Wen Shih
I cut off half e qn. (Ans is in bold). I didnt know I could use that. Thanks!
Originally posted by eagle:tip:
1 way to 'play cheat' is to work backwards... So you sum up the partial fractions on the right hand side, then observe how it was done
Or you try expanding the right hand side partial fractions, then expand left hand side, then observe if there's any pattern linking them
LOL.
Verify that (r+1)² r² (2r+1) - r² (r-1)² (2r-1) = 10r^4 2r²
Hence prove that
n
∑ r^4 = 1/30 n(n+1) (2n+1) (3n² + 3n + 1)
r=1
Note: The result
n
∑ r² = { n(n+1)(2n+1) } / 6
r=1
may be used without proof.
Thanks in advance~!
( already done the part in bold)
omg harder than my sch's tutorial. i dun suppose that you can split r^4 into r^2 x r^2? Think is wrong if you seperately sum.
Hmm.. i would do this qns backwards, from the answer. Really got no idea.
Hi,
The RHS of your identity looks odd. It should have and expression involving r^4 and r^2.
Consider taking the summation of both sides. Then use the method of differences for the LHS, i.e.
(2² 1² 3 - 1² 0² 1)
+ (3² 2² 5 - 2² 1² 3)
+ ...
Try it, thanks!
Cheers,
Wen Shih
Edited. the qn. got a typo.
anyone can further explain method of differences.
like whats the difference for this example:
got
1) (a1 - a2) + (a2 - a3) + ... + (an - an+1) = (a1 - an+1)
2) (a0 - a1) + (a1 - a2) + ... + (an-1 - an) = (a0 - an)
i commonly use 1) but when to use which?
http://4.bp.blogspot.com/_IIevYhD1fT8/Sbyab9hCNOI/AAAAAAAABXU/jV6UQ19ScUM/s1600-h/New+Picture.png
How to solve this equation? i've done the showing part already.
Hi blueftw,
For your first posting, both have the same way of cancelling terms such that only the top-left and bottom-right terms remain. The difference is in the indices the terms take.
The second method of cancelling terms (which you must also remember as it will be tested) will result in having the top-right and bottom-left terms remaining.
To address the question in your second posting, observe that the sum is three times the summation of the expression you have shown. Then, use the method of differences where you will cancel terms via the second method.
Try it, thanks!
P.S. Visit my website www.freewebs.com/weews to learn more about H2 mathematics content. Thanks in advance!
Cheers,
Wen Shih
hey thanks! i managed to solve that question already. have any website that can illustrate the different ways of doing method of differences?
thanks!
Hi,
Questions involving the method of differences are straightforward and there are only two ways to the cancellation of terms. Do not worry :) Thanks!
Cheers,
Wen Shih
thanks WenShih alot for your help (:
hey.
what are the disadvantages of using the stratified sampling method in a US Political Survey. Pop:306,008,000 Sample Size:1,000,000.
Hi,
You may need to find out more about the way the US political survey is carried out. Ask why the population and sample size are given for your consideration also.
This is a question in H2 mathematics? Thanks!
Cheers,
Wen Shih
Hi, I have a Q
Express the general term of the series
2/(3)(5) - 3/(5)(7) + 4/(7)(9) - 5/(9)(11) + ...
in partial fractions (can) and deduce its sum to n terms.
Don't know how to do the 2nd part. Thanks. :)
what's your answer for your partial fractions?
You should be able to do use the method of difference to sum up
Hi,
This is a really good question that is rather unconventional because of its alternating signs. Cambridge has never assessed this type of question in recent years.
You will need to consider two cases:
1. n is even,
2. n is odd.
When n is even, the last row is (-1/4) (1/(2n + 1)) - (-1/4) (1/(2n + 3)).
When n is odd, the last row is ...?
Try it, thanks!
Cheers,
Wen Shih
Don't mind me hijacking topic.
Recurring relations:
The positive numbers Xn satisfy the relation X(n+1) = sqroot Xn + 5 for n = 1, 2, 3...
As n tends to infinity, Xn tends to L.
i) Find the value of L to 3 d.p. and the exact value of L. (DONE)
ii) Prove that (X(n+1) )^ 2 - L^2 = Xn - L (ALSO DONE)
iii) Hence show that if Xn > L, then Xn > X(n+1) > L.
My problem is in proving that Xn > X(n+1)
The HENCE sounds fishy though. I'll think further.
In the meantime, thanks for any assistance.
Hijack over.
Hi Garrick,
'Hence' always suggests that we consider earlier part(s) of the problem.
To show that X_n > X_(n + 1), consider the graphs of y = x and y = sqrt (x + 5) and the fact that both intersect at x = L.
You also need to show that X_(n + 1) > L.
Try it, thanks!
Cheers,
Wen Shih
I could prove X_(n+1) > L.
How does drawing out the two graphs show anything? The only thing i can show is that sqrt (x+5) = x when x = L. It shows that X_(n+1) = X_n when X_n = L. So when X_n > L, ... how do I continue?
Thanks!
Hi,
When X_n > L, which graph is higher? Thanks!
Cheers,
Wen Shih
OH I SEE. How silly I am, missing out the obvious. So it was so simple after all.
Thanks for the help, Mr Wee!
