# Core 2 -Please could someone help ASAPWatch

This discussion is closed.
#1
Hi

I have tried very hard to do these questions but dont seem to get anywhere. Can someone show me the method step by step in order to get the answers (I know the answers as they are in the back of the book, but i dont understand the method).

1) The rth term of an arithmetic progression is 1 + 4r. Find, in terms of n, the sum of the first n terms of the progression.

2) The sum of the first two terms of an arithmetic progression is 18 and the sum of the first four terms is 52. Find the sum of the first eight terms.

3) The sum of the first twenty terms of an arithmetic progression is 50, and the sum of the next twenty terms is -50. Find the sum of the first hundred terms of the progression.

4) The sum of the first hundred terms of an arithmetic progression with first term a and common difference d is T. The sum of the first 50 odd-numbered terms, i.e. the first,third,fifth,...,ninety-ninth, is 0.5T - 1000. Find the value of d.

ALSO .. could someone show me an easy way to work these out:

1) Evaluate the term which is independant of x in the expansion of (x^2 - 1/2x^2)^16.

2) Find the coefficient of x^-12 in the expansion of (x^3-1/x)^24
0
14 years ago
#2
learn the equations so that you can recite them in your sleep...

a= the first term of the AP (forget GPs for the moment)
n= the number of terms
d= the difference (can be + or -) between the terms

to work out what term n is

n=a+(n-1)d test it, learn it!

to sum the series

sum= (n/2)(2a + (n-1)d) test it, learn it!

you may need both equations to solve a problem, or to apply one of them twice... be prepared to leave an unknown in your equation to be found later.

You will commonly end up with the two equations with 2 unknowns. Shouldn't be a problem.

Back soon

Aitch
0
14 years ago
#3
1.rth term is 1+4r. If r = 1, the first term is 5, if r = 2, the second term is 9.

this gives you a (first term) and, by subtraction, d, the common difference.

use the sum equation (which you now know by heart), leaving n as n, and simplify it if possible.

test it with n=3. the first 3 terms are 5, 9, 13, so if your answer with n=3 produces 27, it may be right.

Ok?

Aitch
0
14 years ago
#4
For q 2, take your sum equation (now so well committed to memory that I don't need to quote it...)

first put n=2 and sum =18 into the equation, and get equation1 (2 unknowns, a and d still there)

then put n=4 and sum =52 into the equation, and get equation2 (2 unknowns, a and d still)

solve equations 1 and 2 simultaneously, to get a and d. then check. You should see that your sequence is

7, 11, 15, 19,... you can see then that your sum to 2 terms and sum to 4 terms are correct.

Then use the formula a THIRD time with your found a and d values and n=8.

You can check it by working out terms 5 to 8 and adding them on.

How are you with this so far?

Try number 3, noting that the sum of the first 20 terms is 50, and the sum of the first 40 terms is 0! Apply the formula twice as before.

Aitch
0
14 years ago
#5
your 2 equations should look like:

50=10(2a+19d)
0=20(2a+39d)

when solved these should give a=4.875 and d=-0.25

Note that d is negative; this is why the larger the sequence is, the smaller the sum is. You start with 4.875, so after about 20 terms, the terms are negative.

Put it on a spreadsheet, replicate down, sum it all, then sum the first 20 terms - you can see it more easily.

Then do the sum formula again with n=100

Aitch
0
14 years ago
#6

This one requires a bit more prior thought

Equation1 is OK

T=(100/2)(2a+99d)

For equation 2, the important thing to grasp is that the common difference will be 2d, since we are summing every other term

[compare 2,4,6,8,10,12,14 here d=2
if you take every other term, 2.6.10... d=4]

So equation 2 is

(T/2)-1000 = (50/2)(2a+(49*2d))

When you rearrange the equations, you should get to 2 equations, both of which contain T and 100a, so if you subtract them, you can work out d, which is, fortunately, all you are asked for.

Aitch
0
14 years ago
#7
(Original post by bluebird)
1) Evaluate the term which is independant of x in the expansion of (x^2 - 1/2x^2)^16.

2) Find the coefficient of x^-12 in the expansion of (x^3-1/x)^24
You should work out the first 3 or 4 terms of the expansion and look at what is happening to the powers of x, because the the term which you want to evaluate is the term in which x disappears.

If you work out the first 4 terms in Q1, you should find that the powers of x are (if I've interpreted your Q correctly as (x^2 - 1/(2x^2)))

32, 28, 24, 20
You can see that the powers of x of the x^2 part are decreasing by 2 as the powers of x of the -1/(2x^2) part are increasing by 2. So when they are equal, the x disappears. This will happen when the power in first part of the expression has reduced from 32 to 16, and the power in the second part has increased from 0 to 16. This should be at term 9. Try it. Work out a few more terms if you can't see the progression, even going up to term 9 if you have time.

Q2 is similar. Work out the first 4 or 5 terms in full to see what's happening, then try to work out which term has the coefficient you're looking for, then work out what that term will be.

Aitch
0
#8
I understand all of that. Thanx alot. Could u just help me on just one more question.

the sequence U1, U2, U3,....., where U1 is a given real number is defined by Un+1=Un^2 - 1

a) describe the behaviour of the sequence for each of the cases U1 =0, U1 = 1 and U1= 2. ( I CAN DO THIS)

b) Given that U2=U1, find exactly the two possible values of U1.
c) Given that U3=U1, show that U1^4-2U1^2 -U1 = 0.

I dont see how part b and c can be done.

Thanks alot
0
14 years ago
#9
b) U2 = U1^2 - 1 = U1
U1^2 - U1 - 1 = 0
U1 = [1 +/- sqrt(1 + 4)] / 2

c) u3 = u2^2 - 1
and u2 = u1^2 -1
substituting:-
u3 = (u1^2 - 1)^2 - 1
u3 = u1^4 - 2u1^2
and u3=u1 so
u1 = u1^4 - 2u1^2
u1^4 - 2u1^2 - u1 = 0
0
13 years ago
#10
(Original post by Aitch)
1.rth term is 1+4r. If r = 1, the first term is 5, if r = 2, the second term is 9.

this gives you a (first term) and, by subtraction, d, the common difference.

use the sum equation (which you now know by heart), leaving n as n, and simplify it if possible.

test it with n=3. the first 3 terms are 5, 9, 13, so if your answer with n=3 produces 27, it may be right.

Ok?

Aitch
Hi Everyone,

I've been stuck on this question for quite a while too. I have one question about it.

it says
The rth term of an arithmetic progression is 1 + 4r
why doesn't this mean a+(n-1)d = 1+4r because... it does say the rth term is 1+4r ???

Thanks in advance for any replies =)
0
13 years ago
#11
(Original post by Saiyan)
Hi Everyone,

I've been stuck on this question for quite a while too. I have one question about it.

it says

why doesn't this mean a+(n-1)d = 1+4r because... it does say the rth term is 1+4r ???

Thanks in advance for any replies =)
it does.

consider:

a+(n-1)d = 1+4r

a = 5, since 4x1 + 1 = 5
d = 4, obviously

so:

5+(n-1)4 = 5 + 4n - 1 = 4n + 1

hence the two equations mean the same thing...
0
13 years ago
#12
(Original post by chewwy)
it does.

consider:

a+(n-1)d = 1+4r

a = 5, since 4x1 + 1 = 5
d = 4, obviously

so:

5+(n-1)4 = 5 + 4n - 1 = 4n + 1

hence the two equations mean the same thing...
Oh right, equating

Thanks Alot, chewwy.
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of the Arts London
MA Performance Design and Practice Open Day Postgraduate
Thu, 24 Jan '19
• Coventry University
Sat, 26 Jan '19
• Brunel University London
Sat, 26 Jan '19

### Poll

Join the discussion

#### Are you chained to your phone?

Yes (105)
19.48%
Yes, but I'm trying to cut back (222)
41.19%
Nope, not that interesting (212)
39.33%