H2 math

Destined2REIGN SG

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)} ]

eagle SG

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

wee_ws SG

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

Destined2REIGN SG

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!

VL SG

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.

Destined2REIGN SG

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)

HollowGrowl SG

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.

wee_ws SG

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

Destined2REIGN SG

Edited. the qn. got a typo.

blueftw SG

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?

blueftw SG

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.

wee_ws SG

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

blueftw SG

hey thanks! i managed to solve that question already. have any website that can illustrate the different ways of doing method of differences?

thanks!

wee_ws SG

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

blueftw SG

thanks WenShih alot for your help (:

blueftw SG

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.

wee_ws SG

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

Occam's Razor SG

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. :)

eagle SG

what's your answer for your partial fractions?

You should be able to do use the method of difference to sum up

wee_ws SG

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