Hard Further Maths Series Questions

Challenging, exam-style Further Maths Series questions with worked solutions. Stretch yourself on the hardest standard-results, expanding-brackets, combined-sums, changing-limits problems.

standard-resultsexpanding-bracketscombined-sumschanging-limitslinearityforming-an-equation
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Which of the following is equal to ∑r=13nr\sum_{r=1}^{3n}r?
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=13nr\sum_{r=1}^{3n}r

    Identify the summand and the limits of the sum.

  2. Quote the standard result for ∑r\sum r

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    The sum of the first NN integers.

  3. Substitute the standard results

    ∑r=13nr=3n(3n+1)2\sum_{r=1}^{3n}r=\frac{3n\left(3n+1\right)}{2}

    Each standard result is evaluated at the relevant limit.

  4. Expand and collect like terms

    9n22+3n2\frac{9n^{2}}{2}+\frac{3n}{2}

    Multiplying out gives the polynomial form of the answer.

  5. Factorise the result fully

    3n(3n+1)2\frac{3n\left(3n+1\right)}{2}

    The factorised form is the expected exam answer.

  6. Check the closed form when n=1n=1

    n=1:LHS=6,RHS=6n=1:\quad\text{LHS}=6,\quad\text{RHS}=6

    Direct addition of the terms agrees with the closed form.

  7. Check the closed form when n=2n=2

    n=2:LHS=21,RHS=21n=2:\quad\text{LHS}=21,\quad\text{RHS}=21

    Direct addition of the terms agrees with the closed form.

  8. Check the closed form when n=3n=3

    n=3:LHS=45,RHS=45n=3:\quad\text{LHS}=45,\quad\text{RHS}=45

    Direct addition of the terms agrees with the closed form.

  9. Check the closed form when n=4n=4

    n=4:LHS=78,RHS=78n=4:\quad\text{LHS}=78,\quad\text{RHS}=78

    Direct addition of the terms agrees with the closed form.

  10. Check the closed form when n=5n=5

    n=5:LHS=120,RHS=120n=5:\quad\text{LHS}=120,\quad\text{RHS}=120

    Direct addition of the terms agrees with the closed form.

  11. State the number of terms in the sum

    (3n)−(1)+1=3n\left(3n\right)-\left(1\right)+1=3n

    A useful check on the limits of the sum.

  12. Recall the standard result for the sum of the first NN integers

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    This is one of the three standard results quoted in the formula book.

  13. Recall the standard result for the sum of the first NN squares

    ∑r=1Nr2=16N(N+1)(2N+1)\sum_{r=1}^{N}r^{2}=\frac{1}{6}N\left(N+1\right)\left(2N+1\right)

    This is the standard result for ∑r2\sum r^{2}.

  14. Recall the standard result for the sum of the first NN cubes

    ∑r=1Nr3=14N2(N+1)2\sum_{r=1}^{N}r^{3}=\frac{1}{4}N^{2}\left(N+1\right)^{2}

    This is the standard result for ∑r3\sum r^{3}.

  15. Use the linearity of the summation operator

    ∑(af(r)+bg(r))=a∑f(r)+b∑g(r)\sum\left(af(r)+bg(r)\right)=a\sum f(r)+b\sum g(r)

    Constants may be taken outside a sum and sums may be split term by term.

  16. Select the option equal to this value

    ∑r=13nr=3n(3n+1)2\sum_{r=1}^{3n}r=\frac{3n\left(3n+1\right)}{2}

    This is the closed form of the sum in terms of nn.

Answer
3n(3n+1)2\frac{3n\left(3n+1\right)}{2}
Question 2
9 markschallenging
Which of the following is the value of ∑r=625r3\sum_{r=6}^{25}r^{3}?
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=625r3\sum_{r=6}^{25}r^{3}

    Identify the summand and the limits of the sum.

  2. Rewrite with a lower limit of 11

    ∑r=625r3=∑r=125r3−∑r=15r3\sum_{r=6}^{25}r^{3}=\sum_{r=1}^{25}r^{3}-\sum_{r=1}^{5}r^{3}

    The standard results only apply to sums starting at r=1r=1.

  3. Quote the standard result for ∑r3\sum r^{3}

    ∑r=1Nr3=14N2(N+1)2\sum_{r=1}^{N}r^{3}=\frac{1}{4}N^{2}\left(N+1\right)^{2}

    The sum of the first NN cubes.

  4. Substitute the standard results

    ∑r=625r3=[105625]−[225]\sum_{r=6}^{25}r^{3}=\left[105625\right]-\left[225\right]

    Each standard result is evaluated at the relevant limit.

  5. Simplify

    ∑r=625r3=105400\sum_{r=6}^{25}r^{3}=105400

    Collect the terms over a common denominator and factorise.

  6. Simplify to the final form

    105400105400

    This is the fully simplified answer.

  7. Write out the first few terms

    216+343+512+⋯+15625216+343+512+\cdots+15625

    The sum runs from r=6r=6 to r=25r=25.

  8. Count the number of terms

    25−6+1=2025-6+1=20

    A sum from r=ar=a to r=br=b has b−a+1b-a+1 terms.

  9. Recall the standard result for the sum of the first NN integers

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    This is one of the three standard results quoted in the formula book.

  10. Recall the standard result for the sum of the first NN squares

    ∑r=1Nr2=16N(N+1)(2N+1)\sum_{r=1}^{N}r^{2}=\frac{1}{6}N\left(N+1\right)\left(2N+1\right)

    This is the standard result for ∑r2\sum r^{2}.

  11. Recall the standard result for the sum of the first NN cubes

    ∑r=1Nr3=14N2(N+1)2\sum_{r=1}^{N}r^{3}=\frac{1}{4}N^{2}\left(N+1\right)^{2}

    This is the standard result for ∑r3\sum r^{3}.

  12. Use the linearity of the summation operator

    ∑(af(r)+bg(r))=a∑f(r)+b∑g(r)\sum\left(af(r)+bg(r)\right)=a\sum f(r)+b\sum g(r)

    Constants may be taken outside a sum and sums may be split term by term.

  13. Recall the sum of a constant

    ∑r=1Nc=cN\sum_{r=1}^{N}c=cN

    A constant cc summed over NN terms contributes cNcN, not cc.

  14. Note the link between the sum of cubes and the sum of integers

    ∑r=1Nr3=(∑r=1Nr)2\sum_{r=1}^{N}r^{3}=\left(\sum_{r=1}^{N}r\right)^{2}

    The sum of the first NN cubes is the square of the sum of the first NN integers.

  15. Select the option equal to this value

    ∑r=625r3=105400\sum_{r=6}^{25}r^{3}=105400

    This is the value of the sum.

Answer
105400105400
Question 3
9 markschallenging
Which of the following is equal to ∑r=1nr2(r+3)\sum_{r=1}^{n}r^{2}\left(r+3\right)?
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=1nr2(r+3)\sum_{r=1}^{n}r^{2}\left(r+3\right)

    Identify the summand and the limits of the sum.

  2. Expand the summand

    r2(r+3)=r3+3r2r^{2}\left(r+3\right)=r^{3}+3r^{2}

    Multiplying out lets the standard results be applied term by term.

  3. Split the sum using linearity

    ∑r=1nr2(r+3)=∑r=1nr3+3∑r=1nr2\sum_{r=1}^{n}r^{2}\left(r+3\right)=\sum_{r=1}^{n}r^{3}+3\sum_{r=1}^{n}r^{2}

    Constants come outside and the sum splits over the separate powers of rr.

  4. Quote the standard result for ∑r3\sum r^{3}

    ∑r=1Nr3=14N2(N+1)2\sum_{r=1}^{N}r^{3}=\frac{1}{4}N^{2}\left(N+1\right)^{2}

    The sum of the first NN cubes.

  5. Quote the standard result for ∑r2\sum r^{2}

    ∑r=1Nr2=16N(N+1)(2N+1)\sum_{r=1}^{N}r^{2}=\frac{1}{6}N\left(N+1\right)\left(2N+1\right)

    The sum of the first NN squares.

  6. Substitute the standard results

    ∑r=1nr2(r+3)=n2(n+1)24+n(n+1)(2n+1)2\sum_{r=1}^{n}r^{2}\left(r+3\right)=\frac{n^{2}\left(n+1\right)^{2}}{4}+\frac{n\left(n+1\right)\left(2n+1\right)}{2}

    Each standard result is evaluated at the relevant limit.

  7. Simplify

    ∑r=1nr2(r+3)=n(n+1)(n2+5n+2)4\sum_{r=1}^{n}r^{2}\left(r+3\right)=\frac{n\left(n+1\right)\left(n^{2}+5n+2\right)}{4}

    Collect the terms over a common denominator and factorise.

  8. Expand and collect like terms

    n44+3n32+7n24+n2\frac{n^{4}}{4}+\frac{3n^{3}}{2}+\frac{7n^{2}}{4}+\frac{n}{2}

    Multiplying out gives the polynomial form of the answer.

  9. Factorise the result fully

    n(n+1)(n2+5n+2)4\frac{n\left(n+1\right)\left(n^{2}+5n+2\right)}{4}

    The factorised form is the expected exam answer.

  10. Check the closed form when n=1n=1

    n=1:LHS=4,RHS=4n=1:\quad\text{LHS}=4,\quad\text{RHS}=4

    Direct addition of the terms agrees with the closed form.

  11. Check the closed form when n=2n=2

    n=2:LHS=24,RHS=24n=2:\quad\text{LHS}=24,\quad\text{RHS}=24

    Direct addition of the terms agrees with the closed form.

  12. Check the closed form when n=3n=3

    n=3:LHS=78,RHS=78n=3:\quad\text{LHS}=78,\quad\text{RHS}=78

    Direct addition of the terms agrees with the closed form.

  13. Check the closed form when n=4n=4

    n=4:LHS=190,RHS=190n=4:\quad\text{LHS}=190,\quad\text{RHS}=190

    Direct addition of the terms agrees with the closed form.

  14. Check the closed form when n=5n=5

    n=5:LHS=390,RHS=390n=5:\quad\text{LHS}=390,\quad\text{RHS}=390

    Direct addition of the terms agrees with the closed form.

  15. State the number of terms in the sum

    (n)−(1)+1=n\left(n\right)-\left(1\right)+1=n

    A useful check on the limits of the sum.

  16. Recall the standard result for the sum of the first NN integers

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    This is one of the three standard results quoted in the formula book.

  17. Select the option equal to this value

    ∑r=1nr2(r+3)=n(n+1)(n2+5n+2)4\sum_{r=1}^{n}r^{2}\left(r+3\right)=\frac{n\left(n+1\right)\left(n^{2}+5n+2\right)}{4}

    This is the closed form of the sum in terms of nn.

Answer
n(n+1)(n2+5n+2)4\frac{n\left(n+1\right)\left(n^{2}+5n+2\right)}{4}
Question 4
9 markschallenging
Which of the following is equal to ∑r=2n+14nr\sum_{r=2n+1}^{4n}r?
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=2n+14nr\sum_{r=2n+1}^{4n}r

    Identify the summand and the limits of the sum.

  2. Rewrite with a lower limit of 11

    ∑r=2n+14nr=∑r=14nr−∑r=12nr\sum_{r=2n+1}^{4n}r=\sum_{r=1}^{4n}r-\sum_{r=1}^{2n}r

    The standard results only apply to sums starting at r=1r=1.

  3. Quote the standard result for ∑r\sum r

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    The sum of the first NN integers.

  4. Substitute the standard results

    ∑r=2n+14nr=[2n(4n+1)]−[n(2n+1)]\sum_{r=2n+1}^{4n}r=\left[2n\left(4n+1\right)\right]-\left[n\left(2n+1\right)\right]

    Each standard result is evaluated at the relevant limit.

  5. Simplify

    ∑r=2n+14nr=n(6n+1)\sum_{r=2n+1}^{4n}r=n\left(6n+1\right)

    Collect the terms over a common denominator and factorise.

  6. Expand and collect like terms

    6n2+n6n^{2}+n

    Multiplying out gives the polynomial form of the answer.

  7. Factorise the result fully

    n(6n+1)n\left(6n+1\right)

    The factorised form is the expected exam answer.

  8. Check the closed form when n=1n=1

    n=1:LHS=7,RHS=7n=1:\quad\text{LHS}=7,\quad\text{RHS}=7

    Direct addition of the terms agrees with the closed form.

  9. Check the closed form when n=2n=2

    n=2:LHS=26,RHS=26n=2:\quad\text{LHS}=26,\quad\text{RHS}=26

    Direct addition of the terms agrees with the closed form.

  10. Check the closed form when n=3n=3

    n=3:LHS=57,RHS=57n=3:\quad\text{LHS}=57,\quad\text{RHS}=57

    Direct addition of the terms agrees with the closed form.

  11. Check the closed form when n=4n=4

    n=4:LHS=100,RHS=100n=4:\quad\text{LHS}=100,\quad\text{RHS}=100

    Direct addition of the terms agrees with the closed form.

  12. Check the closed form when n=5n=5

    n=5:LHS=155,RHS=155n=5:\quad\text{LHS}=155,\quad\text{RHS}=155

    Direct addition of the terms agrees with the closed form.

  13. State the number of terms in the sum

    (4n)−(2n+1)+1=2n\left(4n\right)-\left(2n+1\right)+1=2n

    A useful check on the limits of the sum.

  14. Recall the standard result for the sum of the first NN integers

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    This is one of the three standard results quoted in the formula book.

  15. Recall the standard result for the sum of the first NN squares

    ∑r=1Nr2=16N(N+1)(2N+1)\sum_{r=1}^{N}r^{2}=\frac{1}{6}N\left(N+1\right)\left(2N+1\right)

    This is the standard result for ∑r2\sum r^{2}.

  16. Select the option equal to this value

    ∑r=2n+14nr=n(6n+1)\sum_{r=2n+1}^{4n}r=n\left(6n+1\right)

    This is the closed form of the sum in terms of nn.

Answer
n(6n+1)n\left(6n+1\right)
Question 5
9 markschallenging
Find ∑r=1n(2r)3\sum_{r=1}^{n}\left(2r\right)^{3}, giving your answer in a fully factorised form in terms of nn.
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=1n(2r)3\sum_{r=1}^{n}\left(2r\right)^{3}

    Identify the summand and the limits of the sum.

  2. Split the sum using linearity

    ∑r=1n(2r)3=8∑r=1nr3\sum_{r=1}^{n}\left(2r\right)^{3}=8\sum_{r=1}^{n}r^{3}

    Constants come outside and the sum splits over the separate powers of rr.

  3. Quote the standard result for ∑r3\sum r^{3}

    ∑r=1Nr3=14N2(N+1)2\sum_{r=1}^{N}r^{3}=\frac{1}{4}N^{2}\left(N+1\right)^{2}

    The sum of the first NN cubes.

  4. Substitute the standard results

    ∑r=1n(2r)3=2n2(n+1)2\sum_{r=1}^{n}\left(2r\right)^{3}=2n^{2}\left(n+1\right)^{2}

    Each standard result is evaluated at the relevant limit.

  5. Expand and collect like terms

    2n4+4n3+2n22n^{4}+4n^{3}+2n^{2}

    Multiplying out gives the polynomial form of the answer.

  6. Factorise the result fully

    2n2(n+1)22n^{2}\left(n+1\right)^{2}

    The factorised form is the expected exam answer.

  7. Check the closed form when n=1n=1

    n=1:LHS=8,RHS=8n=1:\quad\text{LHS}=8,\quad\text{RHS}=8

    Direct addition of the terms agrees with the closed form.

  8. Check the closed form when n=2n=2

    n=2:LHS=72,RHS=72n=2:\quad\text{LHS}=72,\quad\text{RHS}=72

    Direct addition of the terms agrees with the closed form.

  9. Check the closed form when n=3n=3

    n=3:LHS=288,RHS=288n=3:\quad\text{LHS}=288,\quad\text{RHS}=288

    Direct addition of the terms agrees with the closed form.

  10. Check the closed form when n=4n=4

    n=4:LHS=800,RHS=800n=4:\quad\text{LHS}=800,\quad\text{RHS}=800

    Direct addition of the terms agrees with the closed form.

  11. Check the closed form when n=5n=5

    n=5:LHS=1800,RHS=1800n=5:\quad\text{LHS}=1800,\quad\text{RHS}=1800

    Direct addition of the terms agrees with the closed form.

  12. State the number of terms in the sum

    (n)−(1)+1=n\left(n\right)-\left(1\right)+1=n

    A useful check on the limits of the sum.

  13. Recall the standard result for the sum of the first NN integers

    ∑r=1Nr=12N(N+1)\sum_{r=1}^{N}r=\frac{1}{2}N\left(N+1\right)

    This is one of the three standard results quoted in the formula book.

  14. Recall the standard result for the sum of the first NN squares

    ∑r=1Nr2=16N(N+1)(2N+1)\sum_{r=1}^{N}r^{2}=\frac{1}{6}N\left(N+1\right)\left(2N+1\right)

    This is the standard result for ∑r2\sum r^{2}.

  15. State the final answer

    ∑r=1n(2r)3=2n2(n+1)2\sum_{r=1}^{n}\left(2r\right)^{3}=2n^{2}\left(n+1\right)^{2}

    This is the closed form of the sum in terms of nn.

Answer
2n2(n+1)22n^{2}\left(n+1\right)^{2}

Unlock 29 more Series questions

Create a free account to work through every Further Maths Series question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Series practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Argand diagramsComplex numbers: arithmeticModelling with differential equationsComplex numbers: de MoivreFirst-order differential equationsFurther calculus