Hard Further Maths Series: method of differences Questions

Challenging, exam-style Further Maths Series: method of differences questions with worked solutions. Stretch yourself on the hardest method-of-differences, telescoping, sum-to-infinity, limits problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The function ff is defined by f(r)=1r(r+1)f(r)=\frac{1}{r\left(r+1\right)}. Show that f(r)−f(r+1)=2r(r+1)(r+2)f(r)-f(r+1)=\frac{2}{r\left(r+1\right)\left(r+2\right)} and hence find the value of nn for which ∑r=1n2r(r+1)(r+2)=65132\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{65}{132}.
Show worked solution

Worked solution

  1. State the function to be used

    f(r)=1r(r+1)f(r)=\frac{1}{r\left(r+1\right)}

    The general term will be written as a difference of values of ff.

  2. Form the difference f(r)−f(r+1)f(r)-f(r+1)

    f(r)−f(r+1)=1r(r+1)−1(r+1)(r+2)f(r)-f(r+1)=\frac{1}{r\left(r+1\right)}-\frac{1}{\left(r+1\right)\left(r+2\right)}

    This is the difference that will telescope.

  3. Put the difference over a common denominator

    f(r)−f(r+1)=2r(r+1)(r+2)f(r)-f(r+1)=\frac{2}{r\left(r+1\right)\left(r+2\right)}

    The two terms combine to give exactly the general term of the series.

  4. State the general term as a difference

    ur=2r(r+1)(r+2)=f(r)−f(r+1)withf(r)=1r(r+1)u_{r}=\frac{2}{r\left(r+1\right)\left(r+2\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{r\left(r+1\right)}

    This is the form required for the method of differences.

  5. Write the sum as a difference of two shifted sums

    ∑r=1n2r(r+1)(r+2)=∑r=1nf(r)−∑r=2n+1f(r)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\sum_{r=1}^{n}f(r)-\sum_{r=2}^{n+1}f(r)

    Re-indexing the second sum shows which terms are common to both.

  6. Write the term when r=1r=1

    f(1)−f(2)=12−16f\left(1\right)-f\left(2\right)=\frac{1}{2}-\frac{1}{6}

    Each term is the difference of two values of ff.

  7. Write the term when r=2r=2

    f(2)−f(3)=16−112f\left(2\right)-f\left(3\right)=\frac{1}{6}-\frac{1}{12}

    Each term is the difference of two values of ff.

  8. Write the term when r=n−1r=n-1

    f(n−1)−f(n)=1n(n−1)−1n(n+1)f\left(n-1\right)-f\left(n\right)=\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}

    The penultimate term of the sum.

  9. Write the term when r=nr=n

    f(n)−f(n+1)=1n(n+1)−1(n+1)(n+2)f\left(n\right)-f\left(n+1\right)=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}

    The last term of the sum; note the second part involves f(n+1)f\left(n+1\right).

  10. Add the terms and cancel the interior terms

    ∑r=1n2r(r+1)(r+2)=(12−16)+(16−112)+⋯+(1n(n−1)−1n(n+1))+(1n(n+1)−1(n+1)(n+2))\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{12}\right)+\cdots+\left(\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\right)+\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  11. Identify the surviving terms

    ∑r=1n2r(r+1)(r+2)=12−1(n+1)(n+2)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}

    Exactly 1 term survives at each end of the sum.

  12. Simplify to obtain the closed form

    ∑r=1n2r(r+1)(r+2)=n(n+3)2(n+1)(n+2)\sum_{r=1}^{n}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}

    Collecting over a common denominator gives a single fraction.

  13. Set the closed form equal to the given total

    n(n+3)2(n+1)(n+2)=65132\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}=\frac{65}{132}

    This turns the summation into an equation in nn.

  14. Cross-multiply and collect all terms on one side

    n2+3n−130=0n^{2}+3n-130=0

    Clearing the denominator leaves a polynomial equation in nn.

  15. Factorise the polynomial

    (n−10)(n+13)=0\left(n-10\right)\left(n+13\right)=0

    Factorising exposes the roots of the equation.

  16. Check by summing the 10 terms directly

    ∑r=1102r(r+1)(r+2)=65132\sum_{r=1}^{10}\frac{2}{r\left(r+1\right)\left(r+2\right)}=\frac{65}{132}

    Direct addition of the terms reproduces the given total.

  17. State the value of nn

    n=10n=10

    This is the required upper limit.

Answer
n=10n=10
Question 2
9 markschallenging
Given that 2(2r−1)(2r+1)≡12r−1−12r+1\frac{2}{\left(2r-1\right)\left(2r+1\right)}\equiv\frac{1}{2r-1}-\frac{1}{2r+1}, find ∑r=n+12n2(2r−1)(2r+1)\sum_{r=n+1}^{2n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}, giving your answer as a single fraction in terms of nn.
Show worked solution

Worked solution

  1. Write the general term in partial fractions

    ur=2(2r−1)(2r+1)≡A2r−1+B2r+1u_{r}=\frac{2}{\left(2r-1\right)\left(2r+1\right)}\equiv\frac{A}{2r-1}+\frac{B}{2r+1}

    The denominator is a product of two linear factors, so a two-term decomposition exists.

  2. Multiply through by the denominator

    2≡A(2r+1)+B(2r−1)2\equiv A\left(2r+1\right)+B\left(2r-1\right)

    Clearing the fractions gives an identity valid for all rr.

  3. Substitute r=12r=\frac{1}{2} to find AA

    2=A(2)⇒A=12=A\left(2\right)\Rightarrow A=1

    This root of the first factor kills the BB term.

  4. Substitute r=−12r=-\frac{1}{2} to find BB

    2=B(−2)⇒B=−12=B\left(-2\right)\Rightarrow B=-1

    This root of the second factor kills the AA term.

  5. Check the decomposition by recombining

    12r−1−12r+1=2(2r−1)(2r+1)\frac{1}{2r-1}-\frac{1}{2r+1}=\frac{2}{\left(2r-1\right)\left(2r+1\right)}

    Putting the two fractions over a common denominator returns uru_{r}.

  6. State the general term as a difference

    ur=2(2r−1)(2r+1)=f(r)−f(r+1)withf(r)=12r−1u_{r}=\frac{2}{\left(2r-1\right)\left(2r+1\right)}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{1}{2r-1}

    This is the form required for the method of differences.

  7. Write the sum as a difference of two shifted sums

    ∑r=n+12n2(2r−1)(2r+1)=∑r=n+12nf(r)−∑r=n+22n+1f(r)\sum_{r=n+1}^{2n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}=\sum_{r=n+1}^{2n}f(r)-\sum_{r=n+2}^{2n+1}f(r)

    Re-indexing the second sum shows which terms are common to both.

  8. Write the term when r=n+1r=n+1

    f(n+1)−f(n+2)=12n+1−12n+3f\left(n+1\right)-f\left(n+2\right)=\frac{1}{2n+1}-\frac{1}{2n+3}

    Each term is the difference of two values of ff.

  9. Write the term when r=n+2r=n+2

    f(n+2)−f(n+3)=12n+3−12n+5f\left(n+2\right)-f\left(n+3\right)=\frac{1}{2n+3}-\frac{1}{2n+5}

    Each term is the difference of two values of ff.

  10. Write the term when r=n+3r=n+3

    f(n+3)−f(n+4)=12n+5−12n+7f\left(n+3\right)-f\left(n+4\right)=\frac{1}{2n+5}-\frac{1}{2n+7}

    Each term is the difference of two values of ff.

  11. Write the term when r=2n−1r=2n-1

    f(2n−1)−f(2n)=14n−3−14n−1f\left(2n-1\right)-f\left(2n\right)=\frac{1}{4n-3}-\frac{1}{4n-1}

    The penultimate term of the sum.

  12. Write the term when r=2nr=2n

    f(2n)−f(2n+1)=14n−1−14n+1f\left(2n\right)-f\left(2n+1\right)=\frac{1}{4n-1}-\frac{1}{4n+1}

    The last term of the sum; note the second part involves f(2n+1)f\left(2n+1\right).

  13. Add the terms and cancel the interior terms

    ∑r=n+12n2(2r−1)(2r+1)=(12n+1−12n+3)+(12n+3−12n+5)+⋯+(14n−3−14n−1)+(14n−1−14n+1)\sum_{r=n+1}^{2n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}=\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)+\left(\frac{1}{2n+3}-\frac{1}{2n+5}\right)+\cdots+\left(\frac{1}{4n-3}-\frac{1}{4n-1}\right)+\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)

    Every value of ff between f(n+2)f\left(n+2\right) and f(2n)f\left(2n\right) appears once positively and once negatively, so it cancels.

  14. Identify the surviving terms

    ∑r=n+12n2(2r−1)(2r+1)=12n+1−14n+1\sum_{r=n+1}^{2n}\frac{2}{\left(2r-1\right)\left(2r+1\right)}=\frac{1}{2n+1}-\frac{1}{4n+1}

    Exactly 1 term survives at each end of the sum.

  15. Simplify to obtain the closed form

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

    Collecting over a common denominator gives a single fraction.

Answer
2n(2n+1)(4n+1)\frac{2n}{\left(2n+1\right)\left(4n+1\right)}
Question 3
9 markschallenging
The function ff is defined by f(r)=1r2f(r)=\frac{1}{r^{2}}, and f(r)−f(r+2)=4r+4r2(r+2)2f(r)-f(r+2)=\frac{4r+4}{r^{2}\left(r+2\right)^{2}}. Which of the following is equal to ∑r=1n4r+4r2(r+2)2\sum_{r=1}^{n}\frac{4r+4}{r^{2}\left(r+2\right)^{2}}?
Show worked solution

Worked solution

  1. State the function to be used

    f(r)=1r2f(r)=\frac{1}{r^{2}}

    The general term will be written as a difference of values of ff.

  2. Form the difference f(r)−f(r+2)f(r)-f(r+2)

    f(r)−f(r+2)=1r2−1(r+2)2f(r)-f(r+2)=\frac{1}{r^{2}}-\frac{1}{\left(r+2\right)^{2}}

    This is the difference that will telescope.

  3. Put the difference over a common denominator

    f(r)−f(r+2)=4r+4r2(r+2)2f(r)-f(r+2)=\frac{4r+4}{r^{2}\left(r+2\right)^{2}}

    The two terms combine to give exactly the general term of the series.

  4. State the general term as a difference

    ur=4r+4r2(r+2)2=f(r)−f(r+2)withf(r)=1r2u_{r}=\frac{4r+4}{r^{2}\left(r+2\right)^{2}}=f(r)-f(r+2)\quad\text{with}\quad f(r)=\frac{1}{r^{2}}

    This is the form required for the method of differences.

  5. Write the sum as a difference of two shifted sums

    ∑r=1n4r+4r2(r+2)2=∑r=1nf(r)−∑r=3n+2f(r)\sum_{r=1}^{n}\frac{4r+4}{r^{2}\left(r+2\right)^{2}}=\sum_{r=1}^{n}f(r)-\sum_{r=3}^{n+2}f(r)

    Re-indexing the second sum shows which terms are common to both.

  6. Write the term when r=1r=1

    f(1)−f(3)=1−19f\left(1\right)-f\left(3\right)=1-\frac{1}{9}

    Each term is the difference of two values of ff.

  7. Write the term when r=2r=2

    f(2)−f(4)=14−116f\left(2\right)-f\left(4\right)=\frac{1}{4}-\frac{1}{16}

    Each term is the difference of two values of ff.

  8. Write the term when r=3r=3

    f(3)−f(5)=19−125f\left(3\right)-f\left(5\right)=\frac{1}{9}-\frac{1}{25}

    Each term is the difference of two values of ff.

  9. Write the term when r=4r=4

    f(4)−f(6)=116−136f\left(4\right)-f\left(6\right)=\frac{1}{16}-\frac{1}{36}

    Each term is the difference of two values of ff.

  10. Write the term when r=n−1r=n-1

    f(n−1)−f(n+1)=1(n−1)2−1(n+1)2f\left(n-1\right)-f\left(n+1\right)=\frac{1}{\left(n-1\right)^{2}}-\frac{1}{\left(n+1\right)^{2}}

    The penultimate term of the sum.

  11. Write the term when r=nr=n

    f(n)−f(n+2)=1n2−1(n+2)2f\left(n\right)-f\left(n+2\right)=\frac{1}{n^{2}}-\frac{1}{\left(n+2\right)^{2}}

    The last term of the sum; note the second part involves f(n+2)f\left(n+2\right).

  12. Add the terms and cancel the interior terms

    ∑r=1n4r+4r2(r+2)2=(1−19)+(14−116)+⋯+(1(n−1)2−1(n+1)2)+(1n2−1(n+2)2)\sum_{r=1}^{n}\frac{4r+4}{r^{2}\left(r+2\right)^{2}}=\left(1-\frac{1}{9}\right)+\left(\frac{1}{4}-\frac{1}{16}\right)+\cdots+\left(\frac{1}{\left(n-1\right)^{2}}-\frac{1}{\left(n+1\right)^{2}}\right)+\left(\frac{1}{n^{2}}-\frac{1}{\left(n+2\right)^{2}}\right)

    Every value of ff between f(3)f\left(3\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  13. Identify the surviving terms

    ∑r=1n4r+4r2(r+2)2=(1+14)−(1(n+1)2+1(n+2)2)\sum_{r=1}^{n}\frac{4r+4}{r^{2}\left(r+2\right)^{2}}=\left(1+\frac{1}{4}\right)-\left(\frac{1}{\left(n+1\right)^{2}}+\frac{1}{\left(n+2\right)^{2}}\right)

    Exactly 2 terms survive at each end of the sum.

  14. Note where the boundary sits

    the tail term is f(n+2), not f(n)\text{the tail term is }f\left(n+2\right)\text{, not }f\left(n\right)

    Using f(n)f\left(n\right) here is the classic off-by-one error.

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

    n=1:LHS=89,RHS=89n=1:\quad\text{LHS}=\frac{8}{9},\quad\text{RHS}=\frac{8}{9}

    Adding the 1 term directly agrees with the closed form.

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

    n=2:LHS=155144,RHS=155144n=2:\quad\text{LHS}=\frac{155}{144},\quad\text{RHS}=\frac{155}{144}

    Adding the 2 terms directly agrees with the closed form.

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

    n=3:LHS=459400,RHS=459400n=3:\quad\text{LHS}=\frac{459}{400},\quad\text{RHS}=\frac{459}{400}

    Adding the 3 terms directly agrees with the closed form.

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

    n=5:LHS=530441,RHS=530441n=5:\quad\text{LHS}=\frac{530}{441},\quad\text{RHS}=\frac{530}{441}

    Adding the 5 terms directly agrees with the closed form.

  19. Select the option matching this closed form

    ∑r=1n4r+4r2(r+2)2=n(n+3)(5n2+15n+12)4(n+1)2(n+2)2\sum_{r=1}^{n}\frac{4r+4}{r^{2}\left(r+2\right)^{2}}=\frac{n\left(n+3\right)\left(5n^{2}+15n+12\right)}{4\left(n+1\right)^{2}\left(n+2\right)^{2}}

    This is the closed form of the sum.

Answer
n(n+3)(5n2+15n+12)4(n+1)2(n+2)2\frac{n\left(n+3\right)\left(5n^{2}+15n+12\right)}{4\left(n+1\right)^{2}\left(n+2\right)^{2}}
Question 4
9 markschallenging
The function ff is defined by f(r)=31−r(2r+1)4f(r)=\frac{3^{1-r}\left(2r+1\right)}{4}. Show that f(r)−f(r+1)=r3rf(r)-f(r+1)=\frac{r}{3^{r}} and hence show that the series converges and find ∑r=1∞r3r\sum_{r=1}^{\infty}\frac{r}{3^{r}}.
Show worked solution

Worked solution

  1. State the function to be used

    f(r)=31−r(2r+1)4f(r)=\frac{3^{1-r}\left(2r+1\right)}{4}

    The general term will be written as a difference of values of ff.

  2. Form the difference f(r)−f(r+1)f(r)-f(r+1)

    f(r)−f(r+1)=31−r(2r+1)4−2r+34⋅3rf(r)-f(r+1)=\frac{3^{1-r}\left(2r+1\right)}{4}-\frac{2r+3}{4\cdot3^{r}}

    This is the difference that will telescope.

  3. Put the difference over a common denominator

    f(r)−f(r+1)=r3rf(r)-f(r+1)=\frac{r}{3^{r}}

    The two terms combine to give exactly the general term of the series.

  4. State the general term as a difference

    ur=r3r=f(r)−f(r+1)withf(r)=31−r(2r+1)4u_{r}=\frac{r}{3^{r}}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{3^{1-r}\left(2r+1\right)}{4}

    This is the form required for the method of differences.

  5. Write the sum as a difference of two shifted sums

    ∑r=1nr3r=∑r=1nf(r)−∑r=2n+1f(r)\sum_{r=1}^{n}\frac{r}{3^{r}}=\sum_{r=1}^{n}f(r)-\sum_{r=2}^{n+1}f(r)

    Re-indexing the second sum shows which terms are common to both.

  6. Write the term when r=1r=1

    f(1)−f(2)=34−512f\left(1\right)-f\left(2\right)=\frac{3}{4}-\frac{5}{12}

    Each term is the difference of two values of ff.

  7. Write the term when r=2r=2

    f(2)−f(3)=512−736f\left(2\right)-f\left(3\right)=\frac{5}{12}-\frac{7}{36}

    Each term is the difference of two values of ff.

  8. Write the term when r=3r=3

    f(3)−f(4)=736−112f\left(3\right)-f\left(4\right)=\frac{7}{36}-\frac{1}{12}

    Each term is the difference of two values of ff.

  9. Write the term when r=n−1r=n-1

    f(n−1)−f(n)=32−n(2n−1)4−31−n(2n+1)4f\left(n-1\right)-f\left(n\right)=\frac{3^{2-n}\left(2n-1\right)}{4}-\frac{3^{1-n}\left(2n+1\right)}{4}

    The penultimate term of the sum.

  10. Write the term when r=nr=n

    f(n)−f(n+1)=31−n(2n+1)4−2n+34⋅3nf\left(n\right)-f\left(n+1\right)=\frac{3^{1-n}\left(2n+1\right)}{4}-\frac{2n+3}{4\cdot3^{n}}

    The last term of the sum; note the second part involves f(n+1)f\left(n+1\right).

  11. Add the terms and cancel the interior terms

    ∑r=1nr3r=(34−512)+(512−736)+⋯+(32−n(2n−1)4−31−n(2n+1)4)+(31−n(2n+1)4−2n+34⋅3n)\sum_{r=1}^{n}\frac{r}{3^{r}}=\left(\frac{3}{4}-\frac{5}{12}\right)+\left(\frac{5}{12}-\frac{7}{36}\right)+\cdots+\left(\frac{3^{2-n}\left(2n-1\right)}{4}-\frac{3^{1-n}\left(2n+1\right)}{4}\right)+\left(\frac{3^{1-n}\left(2n+1\right)}{4}-\frac{2n+3}{4\cdot3^{n}}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  12. Identify the surviving terms

    ∑r=1nr3r=34−2n+34⋅3n\sum_{r=1}^{n}\frac{r}{3^{r}}=\frac{3}{4}-\frac{2n+3}{4\cdot3^{n}}

    Exactly 1 term survives at each end of the sum.

  13. Simplify to obtain the closed form

    ∑r=1nr3r=3−n(3n+1−2n−3)4\sum_{r=1}^{n}\frac{r}{3^{r}}=\frac{3^{-n}\left(3^{n+1}-2n-3\right)}{4}

    Collecting over a common denominator gives a single fraction.

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

    n=1:LHS=13,RHS=13n=1:\quad\text{LHS}=\frac{1}{3},\quad\text{RHS}=\frac{1}{3}

    Adding the 1 term directly agrees with the closed form.

  15. Take the limit of the tail term f(n+1)f\left(n+1\right)

    lim⁡n→∞2n+34⋅3n=0\lim_{n\to\infty}\frac{2n+3}{4\cdot3^{n}}=0

    The tail term tends to a finite limit, so the series converges.

  16. Write the sum to infinity as a limit

    ∑r=1∞r3r=lim⁡n→∞(3−n(3n+1−2n−3)4)\sum_{r=1}^{\infty}\frac{r}{3^{r}}=\lim_{n\to\infty}\left(\frac{3^{-n}\left(3^{n+1}-2n-3\right)}{4}\right)

    The sum to infinity is the limit of the partial sums.

  17. Combine the head terms with the limits of the tail terms

    S∞=(34)−(0)S_{\infty}=\left(\frac{3}{4}\right)-\left(0\right)

    The surviving head terms remain; each tail term is replaced by its limit.

  18. State the sum to infinity

    ∑r=1∞r3r=34\sum_{r=1}^{\infty}\frac{r}{3^{r}}=\frac{3}{4}

    The limit exists, so the series converges to this value.

Answer
34\frac{3}{4}
Question 5
9 markschallenging
The function ff is defined by f(r)=31−r(2r+1)4f(r)=\frac{3^{1-r}\left(2r+1\right)}{4}. Show that f(r)−f(r+1)=r3rf(r)-f(r+1)=\frac{r}{3^{r}} and hence find ∑r=1nr3r\sum_{r=1}^{n}\frac{r}{3^{r}}, giving your answer in terms of nn.
Show worked solution

Worked solution

  1. State the function to be used

    f(r)=31−r(2r+1)4f(r)=\frac{3^{1-r}\left(2r+1\right)}{4}

    The general term will be written as a difference of values of ff.

  2. Form the difference f(r)−f(r+1)f(r)-f(r+1)

    f(r)−f(r+1)=31−r(2r+1)4−2r+34⋅3rf(r)-f(r+1)=\frac{3^{1-r}\left(2r+1\right)}{4}-\frac{2r+3}{4\cdot3^{r}}

    This is the difference that will telescope.

  3. Put the difference over a common denominator

    f(r)−f(r+1)=r3rf(r)-f(r+1)=\frac{r}{3^{r}}

    The two terms combine to give exactly the general term of the series.

  4. State the general term as a difference

    ur=r3r=f(r)−f(r+1)withf(r)=31−r(2r+1)4u_{r}=\frac{r}{3^{r}}=f(r)-f(r+1)\quad\text{with}\quad f(r)=\frac{3^{1-r}\left(2r+1\right)}{4}

    This is the form required for the method of differences.

  5. Write the sum as a difference of two shifted sums

    ∑r=1nr3r=∑r=1nf(r)−∑r=2n+1f(r)\sum_{r=1}^{n}\frac{r}{3^{r}}=\sum_{r=1}^{n}f(r)-\sum_{r=2}^{n+1}f(r)

    Re-indexing the second sum shows which terms are common to both.

  6. Write the term when r=1r=1

    f(1)−f(2)=34−512f\left(1\right)-f\left(2\right)=\frac{3}{4}-\frac{5}{12}

    Each term is the difference of two values of ff.

  7. Write the term when r=2r=2

    f(2)−f(3)=512−736f\left(2\right)-f\left(3\right)=\frac{5}{12}-\frac{7}{36}

    Each term is the difference of two values of ff.

  8. Write the term when r=3r=3

    f(3)−f(4)=736−112f\left(3\right)-f\left(4\right)=\frac{7}{36}-\frac{1}{12}

    Each term is the difference of two values of ff.

  9. Write the term when r=n−1r=n-1

    f(n−1)−f(n)=32−n(2n−1)4−31−n(2n+1)4f\left(n-1\right)-f\left(n\right)=\frac{3^{2-n}\left(2n-1\right)}{4}-\frac{3^{1-n}\left(2n+1\right)}{4}

    The penultimate term of the sum.

  10. Write the term when r=nr=n

    f(n)−f(n+1)=31−n(2n+1)4−2n+34⋅3nf\left(n\right)-f\left(n+1\right)=\frac{3^{1-n}\left(2n+1\right)}{4}-\frac{2n+3}{4\cdot3^{n}}

    The last term of the sum; note the second part involves f(n+1)f\left(n+1\right).

  11. Add the terms and cancel the interior terms

    ∑r=1nr3r=(34−512)+(512−736)+⋯+(32−n(2n−1)4−31−n(2n+1)4)+(31−n(2n+1)4−2n+34⋅3n)\sum_{r=1}^{n}\frac{r}{3^{r}}=\left(\frac{3}{4}-\frac{5}{12}\right)+\left(\frac{5}{12}-\frac{7}{36}\right)+\cdots+\left(\frac{3^{2-n}\left(2n-1\right)}{4}-\frac{3^{1-n}\left(2n+1\right)}{4}\right)+\left(\frac{3^{1-n}\left(2n+1\right)}{4}-\frac{2n+3}{4\cdot3^{n}}\right)

    Every value of ff between f(2)f\left(2\right) and f(n)f\left(n\right) appears once positively and once negatively, so it cancels.

  12. Identify the surviving terms

    ∑r=1nr3r=34−2n+34⋅3n\sum_{r=1}^{n}\frac{r}{3^{r}}=\frac{3}{4}-\frac{2n+3}{4\cdot3^{n}}

    Exactly 1 term survives at each end of the sum.

  13. Note where the boundary sits

    the tail term is f(n+1), not f(n)\text{the tail term is }f\left(n+1\right)\text{, not }f\left(n\right)

    Using f(n)f\left(n\right) here is the classic off-by-one error.

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

    n=1:LHS=13,RHS=13n=1:\quad\text{LHS}=\frac{1}{3},\quad\text{RHS}=\frac{1}{3}

    Adding the 1 term directly agrees with the closed form.

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

    n=2:LHS=59,RHS=59n=2:\quad\text{LHS}=\frac{5}{9},\quad\text{RHS}=\frac{5}{9}

    Adding the 2 terms directly agrees with the closed form.

  16. Simplify to obtain the closed form

    ∑r=1nr3r=3−n(3n+1−2n−3)4\sum_{r=1}^{n}\frac{r}{3^{r}}=\frac{3^{-n}\left(3^{n+1}-2n-3\right)}{4}

    Collecting over a common denominator gives a single fraction.

Answer
3−n(3n+1−2n−3)4\frac{3^{-n}\left(3^{n+1}-2n-3\right)}{4}

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