Further Maths Series Practice Questions

Free Further Maths Series practice questions with full step-by-step worked solutions. Covers standard-results, sum-of-integers, sum-of-squares, sum-of-cubes. Practise exam-style problems and check your method.

standard-resultssum-of-integerssum-of-squaressum-of-cubeslinearitycombined-sums
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Evaluate ∑r=110r\sum_{r=1}^{10}r.
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=110r\sum_{r=1}^{10}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. State the final answer

    ∑r=110r=55\sum_{r=1}^{10}r=55

    This is the value of the sum.

Answer
5555
Question 2
2 markseasy
Which of the following is equal to ∑r=1nr\sum_{r=1}^{n}r?
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=1nr\sum_{r=1}^{n}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. Expand and collect like terms

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

    Multiplying out gives the polynomial form of the answer.

  4. Select the option equal to this value

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

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

Answer
n(n+1)2\frac{n\left(n+1\right)}{2}
Question 3
4 marksintermediate
Which of the following is equal to ∑r=1n(5r−2)\sum_{r=1}^{n}\left(5r-2\right)?
Show worked solution

Worked solution

  1. Write the sum in sigma notation

    ∑r=1n(5r−2)\sum_{r=1}^{n}\left(5r-2\right)

    Identify the summand and the limits of the sum.

  2. Split the sum using linearity

    ∑r=1n(5r−2)=5∑r=1nr−2∑r=1n1\sum_{r=1}^{n}\left(5r-2\right)=5\sum_{r=1}^{n}r-2\sum_{r=1}^{n}1

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

  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. Quote the sum of a constant term

    ∑r=1Nc=cN\sum_{r=1}^{N}c=cN

    The constant is added once for every term of the sum.

  5. Substitute the standard results

    ∑r=1n(5r−2)=5n(n+1)2−2n\sum_{r=1}^{n}\left(5r-2\right)=\frac{5n\left(n+1\right)}{2}-2n

    Each standard result is evaluated at the relevant limit.

  6. Select the option equal to this value

    ∑r=1n(5r−2)=n(5n+1)2\sum_{r=1}^{n}\left(5r-2\right)=\frac{n\left(5n+1\right)}{2}

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

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

Worked solution

  1. Write the sum in sigma notation

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

    Identify the summand and the limits of the sum.

  2. Expand the summand

    (2r+1)2=4r2+4r+1\left(2r+1\right)^{2}=4r^{2}+4r+1

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

  3. Split the sum using linearity

    ∑r=1n(2r+1)2=4∑r=1nr2+4∑r=1nr+∑r=1n1\sum_{r=1}^{n}\left(2r+1\right)^{2}=4\sum_{r=1}^{n}r^{2}+4\sum_{r=1}^{n}r+\sum_{r=1}^{n}1

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

  4. 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.

  5. 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.

  6. Quote the sum of a constant term

    ∑r=1Nc=cN\sum_{r=1}^{N}c=cN

    The constant is added once for every term of the sum.

  7. Substitute the standard results

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

    Each standard result is evaluated at the relevant limit.

  8. Simplify

    ∑r=1n(2r+1)2=n(4n2+12n+11)3\sum_{r=1}^{n}\left(2r+1\right)^{2}=\frac{n\left(4n^{2}+12n+11\right)}{3}

    Collect the terms over a common denominator and factorise.

  9. Expand and collect like terms

    4n33+4n2+11n3\frac{4n^{3}}{3}+4n^{2}+\frac{11n}{3}

    Multiplying out gives the polynomial form of the answer.

  10. Factorise the result fully

    n(4n2+12n+11)3\frac{n\left(4n^{2}+12n+11\right)}{3}

    The factorised form is the expected exam answer.

  11. Select the option equal to this value

    ∑r=1n(2r+1)2=n(4n2+12n+11)3\sum_{r=1}^{n}\left(2r+1\right)^{2}=\frac{n\left(4n^{2}+12n+11\right)}{3}

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

Answer
n(4n2+12n+11)3\frac{n\left(4n^{2}+12n+11\right)}{3}
Question 5
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}

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