Part of the Oxford MAT Livestream

These are the solutions for the Sequences and series worksheet. You are encouraged to look at the solutions only after you've had a serious attempt at a question (getting stuck is a normal part of solving problems, and shouldn't be a reason to immediately look at the solutions).

Warm-up

  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.
    • a3=323=6. a10=90.
    • an+1an=((n+1)2(n+1))(n2n)=2n.
    • an+12an+an1=(an+1an)(anan1)=2n2(n1) using the previous part, and this is 2. Alternatively, without using the previous part, an+12an+an1=((n+1)2(n+1))2(n2n)+((n1)2(n1))=2.
  2. A sequence is defined by a0=1 and an=an1+3 for n1. Find a0+a1++a10. Find a1000.
    • This is the sum of the first 11 terms of an arithmetic sequence with first term a=1 and common difference 3. So the sum is 112(2+10×3)=176.
    • a1000=3001.
  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?
    • This is the sum of the first 11 terms of a geometric sequence with first term a=1 and common ratio 13. So the sum is (1311)131=32(1311).
    • a1000=31000.
    • The common ratio is between 1 and 1 so the sum to infinity does converge. The sum to infinity is 1131=32.
  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.
    • In particular, we would need a0=b0 so 1=A+B. Also, we would need a1=b1 so 4=3A+B. So A=32 and B=12.
    • We should check that this works for all n, so let's check that if an1=bn1 then an=bn. We know that an=3an1+1 and we can work out 3bn1+1=3×323n132+1=323n12 which is exactly bn. So if an1=bn1 then an=bn. The sequences match for all n.
  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.
    • At first sight, this doesn't look like enough information; we haven't been told the values of any of the terms in the sequence! The key is that we're asked to give our answer in terms of the first three terms of the sequence without solving for what those are.
    • For example, if we plug in n=0 then we find a0=C. So we've got an expression for C in terms of a0.
    • Now plug in n=1 and n=2 to get a1=A+B+C and a2=4A+2B+C. We want A and B in terms of the variables a0, a1, a2, and we can use the fact that C=a0 to eliminate C. These are simultaneous equations for A and B with solution A=12(a22a1+a0),B=12(a2+4a13a0)
    • Together with C=a0, that gives a solution for A, B, and C in terms of a0, a1, and a2.
    • Compare this with the first question where we calculated an+12an+an1 for a particular sequence of this form.
  6. Simplify 21+22+23++2n for n1.
    • This is the first n terms of a geometric sequence, with sum 2(2n1)/(21)=2n+12.
  7. Simplify 34+35+36++3n for n4.
    • This is the first n3 terms of a geometric sequence, with sum 34(3n31)/(31)=12(3n+134).
  8. When does the sum 1+x3+x6+x9+x12+... converge? Simplify it in the case that it converges.
    • This is a geometric sequence with common ratio x3, and the sum to infinity converges if |x3|<1, which is precisely |x|<1.
    • In that case, it converges to (1x3)1.
  9. When does the sum 2x+x22x34+ converge? Simplify it in the case that it converges.
    • This is a geometric sequence with common ratio x2, and the sum to infinity converges if |x2|<1, which is precisely |x|<2.
    • In that case, it converges to 21+x2=42+x.
  10. Consider the sum of the first n terms of an arithmetic sequence a1, a2, , an with a1=a and a2=a+d. Explain why the sum of the ith term and the (n+1i)th term doesn't depend on i, as long as 1in. By considering separate cases where n is even or where n is odd, deduce that the sum of the first n terms of an arithmetic sequence is n times the average term.
    • The ith term is a+(i1)d and the (n+1i)th term is a+(ni)d. Those sum to 2a+(n1)d, which doesn't depend on i.
    • Case (i) If n is even, then we can pair up the terms like that, pairing the term a1 to an and working inwards. In that case, every pair has the same sum (equal to the first term plus the last term), so the sum of all these terms is n times the average of the first and last terms.
    • Case (ii) If n is odd, then there is a left-over term in the middle when i=(n+1)/2. This value of this term is exactly equal to the average of the first and last terms, so overall the sum is once again the average of the first and last terms multiplied by the number of terms.
  11. Consider the sum of the first n terms of an arithmetic sequence with first term a and constant difference d. Consider the special case d=0. Write down the sum in this case. Now consider the case a=0. In this case, write the sum in terms of the triangle numbers Tn=1+2+3++n=12n(n+1). Hence write down the sum of the first n terms of an arithmetic sequence. Check that this agrees with the formula above.
    • If d=0 then every term is a, so the sum is na.
    • If a=0 then the sum is d(0+1+2++(n1)), which is dTn1=d2(n1)n.
    • The general case is the sum of the two cases above, so the sum is na+d2(n1)n=n2(2a+(n1)d)

MAT questions

MAT 2016 Q1A

  • The first few terms are a1=1, a2=l, a3=l2.
  • This is a geometric sequence, but be careful because we're asked for the product of the terms and not the sum of the terms.
  • We want l1×l2×l3××l15 which is 1×l×l2××l14
  • Using laws of indices this is l1+2++14. So we actually want to add the terms of an arithmetic sequence in the exponent there!
  • 1+2++14=105 using the formula for the sum of the terms of an arithmetic sequence, so the product is l105.
  • The answer is (d).


MAT 2016 Q1G

  • The first few terms are x0=1, x1=x0=1, x2=x0+x1=2, x3=x0+x1+x2=4.
  • I can spot a pattern here; for n1, it looks like xn=2n1. In fact, that's true; each term is double the one before because the sum of all previous terms includes (i) the most recent term (ii) the sum of all the terms before that, and that sum is equal to another copy of the most recent term.
  • So our sequence xn is 1, 1, 2, 4, 8, 16, 32,
  • We're asked for the sum of one-over-each-term. That's 1+1+12+14+18+116+132+
  • I recognise this sum (after the 1+1+ at the start), it's a geometric series.
  • Overall, with the 1+1+ at the start, this adds up to 3.
  • The answer is (d).

 

MAT 2017 Q1C

  • The first few terms are a1=2, a2=6, a3=3, a4=12, a5=16, a6=13, a7=2, a8=6, a9=3, a10=12.
  • This gets into a repeating pattern; a7=a1 and also a8=a2 and so on.
  • So a2017=a2011=a2005=. We should find the remainder when 2017 is divided by 6. This turns out to be 1, so a2017=a1=2.
  • The answer is (d).

 

MAT 2016 Q5

First note that s1=2 because the sum goes as far as n2n, so if n=1 then we go as far as 1×21=2. It's good to know where we're starting!

(i) We're told to set n to some small numbers. Just like in one of the warm-up problems, this gives us simultaneous equations for A and B and C. In this question, we have s1=2 and s2=2+8=10 and s3=2+8+24=34. We also have f(1)=2(A+B)+C, f(2)=4(2A+B)+C, and f(3)=8(3A+B)+C. So
2A+2B+C=28A+4B+C=1024A+8B+C=34

(ii) Now we need to solve those equations; for example we could subtract the first equation from the second, and subtract the second equation from the third, and simplify, to get 3A+B=4 and 4A+B=6. The difference between these equations gives A=2, then substitute this in to find B=2 and C=2.

(iii) Now we're asked to "show that" something, so we'll need to be careful to explain our answer. If sk=f(k) then 2+8+24++k2k=(2k2)2k+2. We'd like to know about sk+1, which is just sk+(k+1)2k+1 because it's the same sum with one extra term at the end. We need to show that this sum is equal to f(k+1).

(iv) Plugging in f(k) for sk and multiplying out all the brackets, I get sk+1=(2k2)2k+2+(k+1)2k+1=k2k+12k+1+2+k2k+1+2k+1. Collecting like terms, this is (2k)2k+1+2. I'll also work out f(k+1)=(2(k+1)2)2k+1+2=(2k)2k+1+2, which matches exactly. So if sk=f(k) then sk+1=f(k+1).

(v) I'm looking for a sum like 2+8+24+ so that I can re-use the work we did for sn. If I ignore the n's in tn, I can see terms like 2(1) and 4(2) and 8(3) which is the sum I'm looking for. So I'm going to try multiplying out the brackets;
tn=n+2n2+4n8+8n24++2n11,

Thinking carefully about what I'm doing to each bracket there, I can also write 2n11 as 2n1(n(n1)) and expand that bracket too! So
tn=n+2n2+4n8+8n24++2n1n2n1(n1).

This is a geometric sequence n+2n+4n+8n++2n1n with sn1 subtracted off. I've got expressions for both of those, so my answer is
tn=(2n1)n(2(n1)2)2n12=2n+1n2

The sum un is just tn divided by 2n with the terms in the opposite order. So un=2n2n21n.

(v) We've got a formula for sk, and we want
nk=1sk=nk=1((2k2)2k+2)=2nk=1(k2k)2nk=1(2k)+nk=02

The last sum is simple, it's just n copies of 2 added together, so it's 2n. The middle sum is the sum of the terms of a geometric sequence. The first sum is exactly the definition of sn, which we've got a formula for! So altogether we've got
nk=1sk=2sk2(2n+11)+2n=2((2n2)2n+2)4(2n1)+2n=n2n+22n+3+8+2n
 

Reflection

  • Spotting a pattern was invaluable in two of the multiple-choice MAT questions. I can only spot a pattern if I work out a few of the terms. How many terms should I work out? I suppose I might keep working out terms until I can see what's going on, or until the sums get too hard.
  • Are there other ways to work out tn? Perhaps we could do something like the previous part of the question; guess a formula and choose constants to make it work.
  • Are there other ways to work out nk=1sk? Perhaps we could expand each sk in terms of the original definition and re-group the terms somehow?
  • Lots of these questions ask us to switch between xk and xn and xn+1 and xn1. Are you happy making that substitution? Given the formula sn=(2n2)2n+2, can you work out sn3 and s2n? (We didn't need either of those to do the question, but you should be able to work them out without too much trouble!)
  • MAT questions might involve more tricky sequences that aren't geometric sequences or arithmetic sequences, like the sequence in the long MAT question above. Even so, we might be able to use what we know about geometric sequences or arithmetic sequences along the way.
Last updated on 15 Aug 2022, 3:11pm. Please contact us with feedback and comments about this page.