Combinations of Series (Edexcel IAL Further Maths)

Spec point — 7.1

Combinations of Series

How do I find combinations of series?

• Use the three formulae:

  • $\sum_{r = 1}^{n} r = 1 \frac{1}{2} n ( n + 1 )$
  • $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n ( n + 1 ) ( 2 n + 1 )$
  • $\sum_{r = 1}^{n} r^{3} = 1 \frac{1}{4} n^{2} ( n + 1 )^{2}$

• Coefficients of $r$, $r^{2}$ and $r^{3}$ can come out of the sum

  • $\sum_{r = 1}^{n} ( a r^{3} + b r^{2} + c r + d ) = a \sum_{r = 1}^{n} r^{3} + b \sum_{r = 1}^{n} r^{2} + c \sum_{r = 1}^{n} r + d n$

    • Constants are multiplied by $n$

  • You may have to expand expressions in $r$

    • $\sum_{r = 1}^{n} ( r + 3 ) ( r + 4 ) = \sum_{r = 1}^{n} ( r^{2} + 7 r + 12 )$

• It is often possible to factorise final answers, in particular taking out:

  • fractional coefficients
  • $n$
  • $( n + 1 )$

• You may be required to write series as the difference of two sums

  • $\sum_{r=5}^{10} r= \sum_{r=1}^{10} r− \sum_{r=1}^{4} r$

Exam Hint Exam questions often refer to these three formulae as the "standard summation formulae".

Worked Example (a) Using standard summation formulae, show that

$\sum_{r = 1}^{n} ( r - 1 ) r ( r + 1 ) = 1 \frac{1}{4} ( n - 1 ) n ( n + 1 ) ( n + 2 )$

Expand and simplify the left-hand side

$\begin{gathered} ( r - 1 ) r ( r + 1 ) & = r ( r^{2} - 1 ) \\ = r^{3} - r \end{gathered}$

Substitute in the formulae $\sum_{r = 1}^{n} r^{3} = 1 \frac{1}{4} n^{2} ( n + 1 )^{2}$ and $\sum_{r = 1}^{n} r = 1 \frac{1}{2} n ( n + 1 )$

$\begin{gathered} \sum_{r = 1}^{n} ( r^{3} - r ) & = \sum_{r = 1}^{n} r^{3} - \sum_{r = 1}^{n} r \\ = 1 \frac{1}{4} n^{2} ( n + 1 )^{2} - 1 \frac{1}{2} n ( n + 1 ) \end{gathered}$* *

Factorise out the quarter, $n$ and $( n + 1 )$ then simplify

$\begin{gathered} \sum_{r = 1}^{n} ( r^{3} - r ) & = 1 \frac{1}{4} n ( n + 1 ) [ n ( n + 1 ) - 2 ] \\ = 1 \frac{1}{4} n ( n + 1 ) ( n^{2} + n - 2 ) \end{gathered}$

Factorise the quadratic on the right-hand side

$\begin{gathered} \sum_{r = 1}^{n} ( r^{3} - r ) & = = 1 \frac{1}{4} n ( n + 1 ) ( n + 2 ) ( n - 1 ) \\ \end{gathered}$

Rearrange into the form asked for by the question

$\begin{gathered} \sum_{r = 1}^{n} ( r - 1 ) r ( r + 1 ) & = 1 \frac{1}{4} ( n - 1 ) n ( n + 1 ) ( n + 2 ) \end{gathered}$

(b) Hence find the value of

$0\times1\times2+1\times2\times3+2\times3\times4+...+20\times21\times22$

It helps to write out the first few terms in the series $\begin{gathered} \sum_{r = 1}^{n} ( r - 1 ) r ( r + 1 ) \end{gathered}$

$\begin{gathered} ( 1 - 1 ) \times 1 \times ( 1 + 1 ) + ( 2 - 1 ) \times 2 \times ( 2 + 1 ) + ( 3 - 1 ) \times 3 \times ( 3 + 1 ) +... \\ = 0 \times 1 \times 2 + 1 \times 2 \times 3 + 2 \times 3 \times 4 +... \end{gathered}$

The question stops at 20 x 21 x 22 This is actually the 21^stterm (not the 20^th term) To find the sum of the first 21 terms, substitute $n=21$ into part (a)

$\begin{gathered} \sum_{r = 1}^{21} ( r - 1 ) r ( r + 1 ) & = 1 \frac{1}{4} ( 21 - 1 ) \times 21 \times ( 21 + 1 ) ( 21 + 2 ) \\ = 1 \frac{1}{4} \times 20 \times 21 \times 22 \times 23 \end{gathered}$

Final answer: 53130

(c) If $\sum_{r = 1}^{n} ( r - 1 ) r ( r + 1 ) = 10 \sum_{r = 1}^{n} r^{2}$, find $n$.

Substitute the answer from part (a) into the left-hand side Substitute the standard formula $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n ( n + 1 ) ( 2 n + 1 )$ into the right-hand side

$\begin{gathered} 1 \frac{1}{4} ( n - 1 ) n ( n + 1 ) ( n + 2 ) & = 10 \times \frac{1}{6} n ( n + 1 ) ( 2 n + 1 ) \end{gathered}$* *

Cancel the terms $n$ and $( n + 1 )$ from both sides It also helps to multiply both sides by 12 to remove fractions

$3 \begin{gathered} ( n - 1 ) ( n + 2 ) \end{gathered} \begin{gathered} = \end{gathered} \begin{gathered} 20 \end{gathered} \begin{gathered} ( 2 n + 1 ) \end{gathered}$

Expand, then form and solve a quadratic in equation

$\begin{gathered} 3 ( n^{2} + n - 2 ) & = 40 n + 20 \\ 3 n^{2} - 37 n - 26 & = 0 \\ ( 3 n + 2 ) ( n - 13 ) & = 0 \\ n = - \frac{2}{3} \text{or} n = 13 \end{gathered}$

$n$ must be a positive integer (as it is used in sigma notation, $\sum_{r = 1}^{n}$)

$n=13$