C2 Sequences and Series Help

OP

Could you explain how you get the answer for b) please

I have:
5(3^n-1) - 5(3^(n-1) - 1)

But Im not really sure of what to do next..

Reply 3

Trangulor

6

Try writing the 3^(n-1) as x(3^n), where x is a number

Reply 4

If I am not missing anything:

Spoiler

edit: too slow

Reply 5

OP

nota bene

If I am not missing anything:

Spoiler

edit: too slow

How about if we take into account that the question says:
"b) Find an expression for the nth term of the series in the form of k(3^n) where k is an exact fraction"

Does that mean that the answer 10(3^(n-1)) is still accepted if k is an exact fraction?

Reply 6

lilydelarocks

How about if we take into account that the question says:
"b) Find an expression for the nth term of the series in the form of k(3^n) where k is an exact fraction"

Does that mean that the answer 10(3^(n-1)) is still accepted if k is an exact fraction?

Oh well, I obviously didn't read the question properly as you need the answer in form of k(3^n) and not k(3^(n-1))... k=10 would not be a problem, but the exponent is, so do as Trangulor suggested instead...

Reply 7

Trangulor

6

Do as I said, and take the 10 into account, writing 10(3^(n-1)) as 10x(3^n), where x is a number

10x will be your exact fraction

Reply 8

OP

Hey guys, need help on another question...

"Show that Sum of r=1 to r=15 (2^r - 12r) = 64094"
(Sorry I can't draw the sign..)

I had an idea of what to do but it doesnt seem to work..Anyways, my first instinct was to use the equation Sn = a(1-r^n)/1-r

So therefore I knew that I had to work out the first term (a) and the ratio (r)

So a = (2^r - 12r)
a = [2^1 - 12(1)]
a = 2 - 12
a = -10

ar = [2² - 12(2)]
ar = (4 - 24)
ar = -20

So the ratio would be -20/-10 = 2

But substituting the values in (assuming that i have the correct numbers) into the equation of Sn = a(1-r^n)/1-r it equals -327670 not 64094 (if I have done everything correctly)

Then I noticed that the ratio wasnt constant:
ar² = [2³ - 12(3)]
ar² = 8 - 36
ar² = -28

So -28/-20 = 1.4 ...which is not the same as 2

Im confused because nothing seems to working out! I may be missing something..or approaching this question in a completly different way..Please help!

Reply 9

steve2005

17

lilydelarocks

!

I suggest you do it in two parts

Picture 27.png9.4KB

Reply 10

ukgea

6

When the ratio isn't constant, it isn't a geometric progression and hence you can't use the formula for the sum of a geometri progression. :wink:

What you can to is to note that you slipt sums up, i.e.

\displaystyle \sum_{r=1}^{15}\left(2^r - 12r\right) = \sum_{r=1}^{15} 2^r - \sum_{r=1}^{15} 12 r

.
You should now be able to evaluate these two sums each on its own.

Reply 11

OP

Got the answer required! Thanks!

Reply 12

steve2005

17

lilydelarocks

Got the answer required! Thanks!

I see you have the answer. For what its worth here is my solution.
NB I made a small typing error in the first post. I put 12 instead of 15

Picture 28.png37.0KB

Reply 13

OP

Yeah I spotted that and figured it was a typo - and I did it the same as you steve2005 except i used the n/2 (a+l) instead of n/2 (2a + (n-1)d).

Thanks again guys..

And just one more question:
"When a ball is dropped onto a horizontal floor it bounces such that it reaches a maximum height of 60% of the height from which it was dropped"

Does this mean the height is increasing by 60% or is the height getting smaller?

Reply 14

If the height from which it was dropped is 1 the maximum height after bounce would be 0.6

Reply 15

OP

Again, Im stuck ..please help!

When a ball is dropped on a horizontal floor it bounces such that it reachs a max height of 60% of the height from which it was dropped

a) find the max height the ball reaches after its fourth bounce when it is initally dropped from 3m above the floor

b) show that when the ball is dropped from a height of h metres above the floor it travels a total distance of 4h metres before coming to rest

I've worked out a) to be 0.648m
but b) im having trouble on..

Reply 16

lilydelarocks

Again, Im stuck ..please help!

When a ball is dropped on a horizontal floor it bounces such that it reachs a max height of 60% of the height from which it was dropped

a) find the max height the ball reaches after its fourth bounce when it is initally dropped from 3m above the floor

b) show that when the ball is dropped from a height of h metres above the floor it travels a total distance of 4h metres before coming to rest

I've worked out a) to be 0.648m
but b) im having trouble on..

I don't agree with you on a) I suggest

3\times0.6^4=0.3888

as it asks for _after_ the 4th bounce...

I am a bit tired, but I'd imagine you could could form the sum

\displaystyle\sum_{k=0}^{n} h+h0.6+h0.6^2+...+h0.6^n

, which you can play around with and hopefully find a suitable sum for answering b :wink:

(There probably is some other, much simpler way to do this, but this is how I'd approach it).

edit: when you have an expression for the sum, then you just take lim n->infinity

edit2: I just did what I suggested above, but it doesn't seem to work out :s-smilie:

, maybe someone can spot a mistake in my working, or reasoning...

Spoiler

Reply 17

Mathmoid

In each bounce after the first one it travels the same distance up as it does down, so the total distance travelled is:

h + 2.(0.6)h + 2.(0.6^2)h + 2.(0.6^3)h + \ldots = 2h\displaystyle\sum_{r=0}^\infty{\left(0.6^r\right)} - h

Reply 18

steve2005

17

lilydelarocks

Again, Im stuck ..please help!

When a ball is dropped on a horizontal floor it bounces such that it reachs a max height of 60% of the height from which it was dropped
b) show that when the ball is dropped from a height of h metres above the floor it travels a total distance of 4h metres before coming to rest
.

Here

Picture 34.png28.7KB

Reply 19

OP