Sequences

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MAT syllabus

Sequences defined iteratively and by formulae. Arithmetic and geometric progressions*. Their sums*. Convergence condition for infinite geometric progressions*.

* Part of full A-level Mathematics syllabus.

 

Revision

  • A sequence an might be defined by a formula for the nth term like an=n2n.
  • A sequence an might be defined with a "recurrence relation" like an+1=f(an) for n0, if we're given the function f(x) and also given a value for the first term a0. (I'm counting from zero, so my "first term" is a0, but it's also common to start with a1, in which case the recurrence relation could be given as an+1=f(an) for n1).
  • The sum of the first n terms of a sequence ak can be written with the notation n1k=0ak (if the first term is a0) or nk=1ak (if the first term is a1).
  • An arithmetic sequence is one where the difference between terms is constant. The terms can be written as a, a+d, a+2d, a+3d, , where a is the first term and d is the common difference.
  • The sum of the first n terms of an arithmetic sequence with first term a and common difference d is n2(2a+(n1)d), which you can remember as "first term plus last term, times the number of terms, divided by two" or n2(a+l), writing l for the last term.
  • A geometric sequence is one where the ratio between consecutive terms is constant. The terms can be written as a, ar, ar2, ar3, where a is the first term and r is the common ratio.
  • The sum of the first n terms of a geometric sequence with first term a and common ratio r is a(1rn)1r. One way to remember this is to remember what happens if we multiply the sum of the first n terms of a geometric series by (1r),
    (1r)(a+ar++arn1)=(aar)+(arar2)++(arn1arn)=aarn.
  • For a geometric sequence an, the sum to infinity is written as k=0ak. If the common ratio r satisfies |r|<1 then this is equal to a1r. If |r|1 then this sum to infinity does not converge (it does not approach any particular real number).
  • A periodic sequence has terms that repeat in a cycle; there is some L such that for all n0, we have an+L=an. The order of a periodic sequence is the minimum such L.

 

Revision Questions

  1. A sequence is defined by an=n2n. What is a3? What is a10? Find an+1an in terms of n. Find an+12an+an1 in terms of n.
  2. A sequence is defined by a0=1 and an=an1+3 for n1. Find a0+a1++a10. Find a1000.
  3. A sequence is defined by a0=1 and an=an13 for n1. Find a0+a1++a10. Find a1000. Does the sum of all the terms of this sequence converge? If it does, what is the sum to infinity?
  4. A sequence is defined by a0=1 and an=3an1+1 for n1. A sequence bn is defined by bn=A×3n+B where A and B are real numbers. Find values for A and B such that an=bn for all n0.
  5. A sequence is defined by an=An2+Bn+C where A, B, and C are real numbers. Find A, B, and C in terms of a0, a1, and a2.    
  6. When does the sum 1+x3+x6+x9+x12+... converge? Simplify it in the case that it converges.
  7. When does the sum 2x+x22x34+ converge? Simplify it in the case that it converges.
  8. If the first term of an arithmetic progression is 5 and the common difference is 3, what is the 15th term?
  9. The sum of the first k terms of an arithmetic progression is equal to the sum of the next k terms. What can you deduce?
  10. If the sum of the first n terms of an arithmetic progression is 3n2+5n, what is the nth term?
  11. What is the sum of the first 100 positive even integers (starting at 2)?
  12. The first term of a geometric progression is 3 and the third term is 27. Find two possibilities for the sum of the first 5 terms.
  13. A sequence is defined by a0=3 and then for n1 an is the sum of all previous terms. Find an in terms of n for n1.
  14. A sequence is defined by C0=1 and then for n0, Cn+1=ni=0CiCni.
    Find C1 and C2 and C3 and C4.

 

MAT questions

MAT 2016 Q1A

A sequence an has first term a1=1, and subsequent terms defined by an+1=lan for n1. What is the product of the first 15 terms of the sequence?

(a) l14,
(b) 15+l14,
(c) 15l14,
(d) l105,
(e) 15+l105.

 

Hints

  • Work out the first few terms of the sequence.
  • What's the product of the first three terms of the sequence? Can you simplify your answer? What sum did you need to do in order to simplify your answer?
  • How would that change with 15 terms? What would the 15th term be?
  • What happens if l=1?

 

MAT 2017 Q1C

A sequence (an) has the property that an+1=anan1 for every n2. Given that a1=2 and a2=6, what is a2017?

(a) 16,
(b) 23,
(c) 32,
(d) 2,
(e) 3.

 

Hints

  • Work out the first few terms of the sequence.
  • You might find that after a while the calculations you're doing repeat previous calculations. Will that keep happening?
  • For which values of n is an=2? You know that n=1 is one such value of n.

 

MAT 2016 Q1G

The sequence xn, where n0, is defined by x0=1 and xn=n1k=0xkfor n1.

The sum k=01xk equals

(a) 1,
(b) 65,
(c) 85,
(d) 3,
(e)  275.

 

Hints

  • Work out the first few terms of the sequence.
  • The Σ notation means that x1=x0, and then x2=x0+x1, and then x3=x0+x1+x2, and so on. Each term in the sequence is the sum of all the previous terms.
  • Now work out 1/xn for the values of xn you've calculated.
  • It's not quite true that this sequence has a common ratio, but it's almost true!
  • The sum of all the terms of the sequence is the same thing as the sum of the first two plus the sum of all the others.

 

MAT 2008 Q2

(i) Find a pair of positive integers, x1 and y1, that solve the equation
(x1)22(y1)2=1.

(ii) Given integers a,b, we define two sequences x1,x2,x3, and y1,y2,y3, by setting
xn+1=3xn+4yn,yn+1=axn+byn,for n1.

Find positive values for a,b such that
(xn+1)22(yn+1)2=(xn)22(yn)2.

(iii) Find a pair of integers X,Y which satisfy X22Y2=1 such that X>Y>50.

(iv) Using the values of a and b found in part (ii), what is the approximate value of xn/yn as n increases?


Hints

(i) Searching small values of x1 or small values of y1 is a good idea here. We're only asked to find a pair, not all such pairs. The question doesn't specific whether zero counts as a positive number (some people do count it, some people don't), so that's up to you.

(ii) Substitute everything in and hope for the best. We want this to be true for lots of different values of xn and yn (presumably), so we might aim to do something like comparing coefficients.

Hopefully this will give us some equations involving a and b. We're not too worried about finding all possible solutions here; we're just looking for anything that works, and that has a and b positive.

(iii) This part of the question is all about understanding the previous part. We found a way to generate a sequence xn and a sequence yn, and we showed that the sequences satisfy some sort of rule. Why did we do that? What's it got to do with the value of X22Y2?

It's easy to get distracted by the relationship that we've just proved if you're looking for a link between xn+1 and xn. Don't forget that we also have rules like xn+1=3xn+4yn which are easier to work with if we want to calculate xn+1 from our knowledge of previous values of xn and yn.

Alternatively, try large numbers Y until you find one with 2Y2+1 equal to a square number. This might take a while!

(iv) From our work on the previous parts, we know that xn and yn satisfy a particular equation. We also know that xn and yn will be large for large n. Can you see how to convert the equation you've got into a fact about xn/yn?


Extension

[Just for fun, not part of the MAT question]

  •  Find some rational approximations to 3 with a similar method.


MAT 2016 Q5

This question concerns the sum sn defined by
sn=2+8+24++n2n. 
(i) Let f(n)=(An+B)2n+C for constants A, B and C yet to be determined, and suppose sn=f(n) for all n1.  By setting n=1,2,3, find equations that must be satisfied by A, B and C.

(ii) Solve the equations from part (i) to obtain values for A, B and C.

(iii) Using these values, show that if sk=f(k) for some k1 then sk+1=f(k+1).

You may now assume that f(n)=sn for all n1.

(iv) Find simplified expressions for the following sums:
    tn=n+2(n1)+4(n2)+8(n3)++2n11,un=12+24+38++n2n.

(v) Find the sum nk=1sk.


Hints

(i) Check that you understand the relationship between the expression n2n and the numbers 2 and 8 and 24. But notice that these aren't the values of sn. The sequence sn involves adding these numbers together, so that when n=2, the value of sn is the sum 2+8=10 (not just 8).

You'll need the values of s1 and s2 and s3, and you'll need to calculate f(1) and f(2) and f(3) in terms of A and B and C.

It's a good idea to write out your work clearly, so that you have three equations in a tidy format, ready for the next part.

(ii) My equations each have +C, so I can take the difference of the first and second, or I can take the difference of the second and third, and either way I'll eliminate C. That gives me two equations for two unknowns (A and B).

(iii) What's the difference between sk+1 and sk? If we knew a neat formula for sk, that would clearly help us calculate sk+1 (you wouldn't just start the sum again). You might even have spotted that shortcut while working out s3 in a previous part. What's the difference in terms of k?

For sk+1 to be equal to f(k+1), we need to show that the change from sk to sk+1 matches the change from f(k) to f(k+1).

We have values for A and B and C from the previous part, so there's no mystery to f(k+1). We could calculate the difference between f(k) and f(k+1) in terms of k.

For all of this to work, we'll need to see matching expressions for those differences.

(iv) For tn, expand each bracket. Collect terms together, and watch out for a sum that we've already done.

For un, find a relationship between tn and un.

(v) Use the formula for sk.

Consider the terms corresponding to A and B and C separately. Those are each just constants (which you know the value for!) and they can be brought outside each sum. For example, nk=1(Ak2k)=Ank=1(k2k).

Once again, you'll need to recognise a sum that you've already done.

 

Extension

[Just for fun, not part of the MAT question]

  •  Find the sum nm=1(mk=1sk).
     
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