Brief Qs (1 Viewer)

[lyounamu]

Show that Ke^0, Ke^(-1), Ke^(-2), .... are the first terms of geometric series.


For this question, I just need to show that r is same throughout the series, yes?

So would this be fair:

For it to be the first terms of geometric series, its r (i.e. Tn/T(n-1)) should be constant throughout the series. (i.e. Tn/T(n-1) = T(n-1)/T(n-2))

i.e. Ke^(-2)/Ke^(-1) = 1/e

And Ke^(-1)/Ke^0 = 1/e.

Since r= Tn/T(n-1) = T(n-1)/T(n-2)=1/e,

they are the first terms of geometric series.

 

[vds700]

Show that Ke^0, Ke^(-1), Ke^(-2), .... are the first terms of geometric series.


For this question, I just need to show that r is same throughout the series, yes?

So would this be fair:

For it to be the first terms of geometric series, its r (i.e. Tn/T(n-1)) should be constant throughout the series. (i.e. Tn/T(n-1) = T(n-1)/T(n-2))

i.e. Ke^(-2)/Ke^(-1) = 1/e

And Ke^(-1)/Ke^0 = 1/e.

Since r= Tn/T(n-1) = T(n-1)/T(n-2)=1/e,

they are the first terms of geometric series.

That looks fine to me.

 

[lyounamu]

That looks fine to me.

Thanks, Andrew!

 

[lyounamu]

When Jill was born her mother deposited $180 into a Trust Account earning 12% P.A.interest compounded annually. She continued to deposit $180 into this account each time Jill had a birthday. The last payment was made on Jill's 16th birthday. Calculate the total amount in the account on Jill's 25th birthday.

In my opinion, the solution is wrong but I am not sure as to whether I am right or not. I got $21340.47847...

Please post up your solution if you got otherwise. Thanks

 

[veloc1ty]

I got $21839.63.

Is that the answer or am I wrong?

 

[lyounamu]

I got $21839.63.

Is that the answer or am I wrong?

Answer is $24400.49

In their working out, they did something wrong...I mean, at least I perceive them to have...

 

[veloc1ty]

I got it by playing with my second last step...

A(1) = 180 + 180*1.12 [i.e. the first year after it is established the initial $180 has been in bank 1 year and then $180 more for the 1st bday]
A(2) = 180*1.12^2 + 180*1.12 + 180
A(3) = 180(1 + 1.12 + 1.12^2 + 1.12^3)
A(16) = 180(1 + 1.12 + ... + 1.12^16)
A(16) = 180{[1(1.12^17 - 1)]/(1.12 - 1)}
A(16) = 8799.06

A(25) = 8799.06 * 1.12^9
A(25) = $24400.49

 

[lyounamu]

Here is my working-out for those who may want to see whether I am wrong or not.

Here we go:

First payment:
180 x 1.12
2nd payment:
180 x 1.12^2 + 180 x 1.12
16th payment:
180 x 1.12 ^16 + 180 x 1.12^15 +...+180

From here, I added them up because the payments stopped here.
S16 = 7695.59047....

BUT we still have some time to go till Jill's 25th birthday (i.e. 9 years)

So... 7695.59047... x 1.12^9 = 21340.47847...

And this is another method that I used to see from a different perspective (but basically same principple):

Using the geometric sum method:

S16 = 180 x 1.12^9 (1.12^16-1)/0.12 = 21340.47847...

 

[veloc1ty]

Yeah you messed up the very first payment. Don't you hate that?

 

[Iruka]

I can see your mistake - you should have seventeen terms to sum, not sixteen, because the mother made the first deposit on the 0th birthday, not the first birthday. In this case, the sum is 8799.061333, not 7695.590476 (according to my computer).

 

[lyounamu]

I can see your mistake - you should have seventeen terms to sum, not sixteen, because the mother made the first deposit on the 0th birthday, not the first birthday. In this case, the sum is 8799.061333, not 7695.590476 (according to my computer).

I thought 1st birthday = birth

 

[veloc1ty]

Assume "16th birthday" to be the day she is 16, not the 16th one including the actual birth day.

 

[lyounamu]

Assume "16th birthday" to be the day she is 16, not the 16th one including the actual birth day.

Yeah, my mistake.

 

[lyounamu]

http://www.boredofstudies.org/courses/maths/2u/2001_Maths2_T_Manly_q.pdf

Please look at Q 5(a).

For that question, do we just find the approximate area for both and subtract each other?


And please quickly do Q4(c). It only takes 5 minutes to complete. See if your answer is 579 for the approximate area. Thanks~

 

[lyounamu]

Is there anyone to look at the questions I queried?

 

[lyounamu]

yep




i get 508. just divide the area into 4 seperate trapeziums and calculate the area of each.

A ~ 10(d1 + d2)/2 + 10(d2 + d3)/2 + 10(d5 + d4)/2 + 10(d5 + d6)/2

I see what you did, thanks. I actually solved in 2 ways:

one way as in the strip
and in the other way that you implemented and I got different answers so I wasn't sure. Thanks for confirming that!!!