Proof by induction Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Proof by induction questions. See exactly how to solve problems on series-evaluate, induction, series, summation.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
It can be proved by induction that ∑r=1n[r]=n(n+1)2\sum_{r=1}^{n}\left[r\right]=\frac{n (n + 1)}{2} for all positive integers nn. Use this result to evaluate ∑r=110[r]\sum_{r=1}^{10}\left[r\right].

Worked solution

  1. Quote the proved closed form

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

    The induction has already established this formula.

  2. Substitute n=10n=10 into the closed form

    n=10:55n=10:\quad 55

    Replace nn by the required upper limit.

  3. State the proposition to be proved

    P(n): ∑r=1n[r]=n(n+1)2P(n):\ \sum_{r=1}^{n}\left[r\right]=\frac{n (n + 1)}{2}

    Name the statement so the base case and inductive step can refer to it.

  4. State the value of the sum

    ∑r=110[r]=55\sum_{r=1}^{10}\left[r\right]=55

    This is the required value of the series.

Answer
5555
Question 2
2 markseasy
It can be proved by induction that ∑r=1n[r2]=n(n+1)(2n+1)6\sum_{r=1}^{n}\left[r^{2}\right]=\frac{n (n + 1) (2 n + 1)}{6} for all positive integers nn. Use this result to evaluate ∑r=16[r2]\sum_{r=1}^{6}\left[r^{2}\right].

Worked solution

  1. Quote the proved closed form

    ∑r=1n[r2]=n(n+1)(2n+1)6\sum_{r=1}^{n}\left[r^{2}\right]=\frac{n (n + 1) (2 n + 1)}{6}

    The induction has already established this formula.

  2. Substitute n=6n=6 into the closed form

    n=6:91n=6:\quad 91

    Replace nn by the required upper limit.

  3. State the proposition to be proved

    P(n): ∑r=1n[r2]=n(n+1)(2n+1)6P(n):\ \sum_{r=1}^{n}\left[r^{2}\right]=\frac{n (n + 1) (2 n + 1)}{6}

    Name the statement so the base case and inductive step can refer to it.

  4. State the value of the sum

    ∑r=16[r2]=91\sum_{r=1}^{6}\left[r^{2}\right]=91

    This is the required value of the series.

Answer
9191
Question 3
2 markseasy
It can be proved by induction that ∑r=1n[2r−1]=n2\sum_{r=1}^{n}\left[2 r - 1\right]=n^{2} for all positive integers nn. Use this result to evaluate ∑r=112[2r−1]\sum_{r=1}^{12}\left[2 r - 1\right].

Worked solution

  1. Quote the proved closed form

    ∑r=1n[2r−1]=n2\sum_{r=1}^{n}\left[2 r - 1\right]=n^{2}

    The induction has already established this formula.

  2. Substitute n=12n=12 into the closed form

    n=12:144n=12:\quad 144

    Replace nn by the required upper limit.

  3. State the proposition to be proved

    P(n): ∑r=1n[2r−1]=n2P(n):\ \sum_{r=1}^{n}\left[2 r - 1\right]=n^{2}

    Name the statement so the base case and inductive step can refer to it.

  4. State the value of the sum

    ∑r=112[2r−1]=144\sum_{r=1}^{12}\left[2 r - 1\right]=144

    This is the required value of the series.

Answer
144144
Question 4
2 markseasy
It can be proved by induction that ∑r=1n[r3]=n2(n+1)24\sum_{r=1}^{n}\left[r^{3}\right]=\frac{n^{2} (n + 1)^{2}}{4} for all positive integers nn. Use this result to evaluate ∑r=15[r3]\sum_{r=1}^{5}\left[r^{3}\right].

Worked solution

  1. Quote the proved closed form

    ∑r=1n[r3]=n2(n+1)24\sum_{r=1}^{n}\left[r^{3}\right]=\frac{n^{2} (n + 1)^{2}}{4}

    The induction has already established this formula.

  2. Substitute n=5n=5 into the closed form

    n=5:225n=5:\quad 225

    Replace nn by the required upper limit.

  3. State the proposition to be proved

    P(n): ∑r=1n[r3]=n2(n+1)24P(n):\ \sum_{r=1}^{n}\left[r^{3}\right]=\frac{n^{2} (n + 1)^{2}}{4}

    Name the statement so the base case and inductive step can refer to it.

  4. State the value of the sum

    ∑r=15[r3]=225\sum_{r=1}^{5}\left[r^{3}\right]=225

    This is the required value of the series.

Answer
225225
Question 5
2 markseasy
To prove by induction that ∑r=1n[r(r+1)]=n(n+1)(n+2)3\sum_{r=1}^{n}\left[r (r + 1)\right]=\frac{n (n + 1) (n + 2)}{3}, the base case must be checked first. Evaluate the left-hand side ∑r=1n[r(r+1)]\sum_{r=1}^{n}\left[r (r + 1)\right] when n=1n=1.

Worked solution

  1. Write the left-hand side with n=1n=1

    ∑r=11[r(r+1)]\sum_{r=1}^{1}\left[r (r + 1)\right]

    The sum has a single term when n=1n=1.

  2. Substitute r=1r=1 into the summand

    u1=2=2u_1=2=2

    Only the r=1r=1 term appears.

  3. State the proposition to be proved

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

    Name the statement so the base case and inductive step can refer to it.

  4. State the base-case value

    ∑r=11[r(r+1)]=2\sum_{r=1}^{1}\left[r (r + 1)\right]=2

    This is the value of the left-hand side when n=1n=1.

Answer
22

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