Standard Series (Edexcel IAL Further Maths)

Sums of Natural Numbers

Spec point — 7.1

What is sigma notation?

• Sigma notation represents sums as follows:$\sum_{r=1}^{5} r=1+2+3+4+5$

  • $r$ counts in integers from the lower limit to the upper limit

• It works for functions of $r$

  • $\sum_{r = 1}^{5} ( 3 r - 1 ) = ( 3 \times 1 - 1 ) + ( 3 \times 2 - 1 ) + ( 3 \times 3 - 1 ) +... + ( 3 \times 5 - 1 )$

• The sum of the first$n$terms has an upper limit of $n$

  • $\sum_{r=1}^{n} r=1+2+3+...+n$

How do I write a series as the difference of two sums?

• When the lower limit is not 1, such as$\sum_{r=5}^{10} r=5+6+7+...+10$, you can write it as the difference between two sums:

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

  • This is because$5 + 6 + 7 + 8 + 9 + 10 = ( 1 + 2 + 3 + 4 +... + 10 ) - ( 1 + 2 + 3 + 4 )$

    • The 1, 2, 3 and 4 cancel out, leaving 5, 6, ..., 10

  • Be careful: the upper limit of the second sum is one less than the lower limit of the original sum!

What properties of sigma notation do I need to know?

• You need to know the following rules:

  • $\sum_{r=1}^{n} k=k+k+...+k=k n$

    • The sum of a constant is the constant$\times n$

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

    • Coefficients of $r$* *can come out of the sum
    • Constants are multiplied by $n$

What is the formula for the sum of the natural numbers?

• $\sum_{r=1}^{n} r=1+2+3+...+n$ is the sum of the natural numbers

  • Natural numbers are positive integers

• It has the standard formula:

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

  • You can work the formula out using arithmetic series

    • first term 1, common difference 1

• You may use the formula in calculations without proof

Exam Hint The formula for the sum of natural numbers is not given in the Formulae Booklet!

Worked Example A sum of $n$ terms is given by $\sum_{r = 1}^{n} ( 2 r + 1 )$.

(a) Using any standard summation formulae, show that $\sum_{r = 1}^{n} ( 2 r + 1 ) = n ( n + A )$ where $A$ is a positive integer to be found.

Use the rule that $\sum_{r = 1}^{n} ( a r + b ) = a \sum_{r = 1}^{n} r + b n$

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

Substitute in the formula $\sum_{r = 1}^{n} r = 1 \frac{1}{2} n ( n + 1 )$* *

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

Expand and simplify

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

Factorise the right-hand side

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

Check it has the right form, $n ( n + A )$

Final answer: $\begin{gathered} \sum_{r = 1}^{n} ( 2 r + 1 ) & = n ( n + 2 ) \end{gathered}$where$\begin{gathered} A = 2 \end{gathered}$

(b) Hence find the sum of the odd numbers from 3 to 81.

It often helps to write out the first few terms and the last term of the series

$\sum_{r = 1}^{n} ( 2 r + 1 ) = ( 2 \times 1 + 1 ) + ( 2 \times 2 + 1 ) + ( 2 \times 3 + 1 ) +... + ( 2 \times n + 1 )$

Simplify these terms

$\sum_{r = 1}^{n} ( 2 r + 1 ) = 3 + 5 + 7 +... + ( 2 n + 1 )$

This is the sum of the odd numbers from 3 to $( 2 n + 1 )$ The question wants the sum of odd numbers from 3 to 81 Use the last term to find $n$

• *$\begin{gathered} 2 n + 1 & = 81 \\ 2 n & = 80 \\ n = 40 \end{gathered}$

The sum of the first 40 terms gives the sum of odd numbers from 3 to 81 Substitute $n=40$ into the formula in part (a)

$40 \times ( 40 + 2 )$

Final answer: 1680

Sums of Squares

Spec point — 7.1

What is the formula for the sum of the squares of the natural numbers?

• $\sum_{r = 1}^{n} r^{2} = 1^{2} + 2^{2} + 3^{2} +... + n^{2} = 1 + 4 + 9 +... + n^{2}$ is the sum of the squares of the natural numbers

• It has it the standard formula:

  • $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n ( n + 1 ) ( 2 n + 1 )$
  • You may use the formula in calculations without proof

• Note that $\sum_{r = 1}^{n} ( a r^{2} + b ) = a \sum_{r = 1}^{n} r^{2} + b n$

  • Coefficients of* *$r^{2}$ can come out of the sum
  • Constants are multiplied by $n$

Exam Hint This formula is given in the Formulae Booklet.

Worked Example (a) Evaluate $\sum_{r = 1}^{30} r^{2}$

Evaluate means "find the value of" This is the sum of the first 30 square numbers Substitute $n=30$ into the formula $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n ( n + 1 ) ( 2 n + 1 )$* *

$\begin{gathered} \sum_{r = 1}^{30} r^{2} & = \frac{1}{6} \times 30 \times ( 30 + 1 ) ( 2 \times 30 + 1 ) \\ = \frac{1}{6} \times 30 \times 31 \times 61 \end{gathered}$* *

Final answer: 9455

(b) Show that $\sum_{r = m + 1}^{2 m} r^{2} = \frac{1}{6} m ( P m + 1 ) ( Q m + 1 )$ where $P<Q$ and where $P$ and $Q$ are prime numbers to be found.

The lower limit is not 1, so write it as the difference between two sums It can help to imagine simpler numbers, $\sum_{r = 5}^{10} r^{2} = \sum_{r = 1}^{10} r^{2} - \sum_{r = 1}^{4} r^{2}$

$\sum_{r = m + 1}^{2 m} r^{2} = \sum_{r = 1}^{2 m} r^{2} - \sum_{r = 1}^{m} r^{2}$

Now use the formula $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n ( n + 1 ) ( 2 n + 1 )$ for each sum on the right

$\sum_{r = m + 1}^{2 m} r^{2} = \frac{1}{6} ( 2 m ) ( 2 m + 1 ) ( 4 m + 1 ) - \frac{1}{6} m ( m + 1 ) ( 2 m + 1 )$

Factorise out$\frac{1}{6}$ , $m$ and $( 2 m + 1 )$ then simplify

$\begin{gathered} \sum_{r = m + 1}^{2 m} r^{2} & = \frac{1}{6} m ( 2 m + 1 ) [ 2 ( 4 m + 1 ) - ( m + 1 ) ] \\ = \frac{1}{6} m ( 2 m + 1 ) ( 7 m + 1 ) \end{gathered}$

Check this is in the right form 2 and 7 are prime numbers and 2 < 7

Final answer: $\begin{gathered} \sum_{r = m + 1}^{2 m} r^{2} & = \frac{1}{6} m ( 2 m + 1 ) ( 7 m + 1 ) \end{gathered}$where$\begin{gathered} P = 2 \end{gathered}$and$\begin{gathered} Q = 7 \end{gathered}$

Sums of Cubes

Spec point — 7.1

What is the formula for the sum of the cubes of the natural numbers?

• $\sum_{r = 1}^{n} r^{3} = 1^{3} + 2^{3} + 3^{3} +... + n^{3} = 1 + 8 + 27 +... + n^{3}$ is the sum of the cubes of the natural numbers

• It has it the standard formula:

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

  • You may use the formula in calculations without proof

  • It is the square of the formula for the sum of the natural numbers

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

• Note that $\sum_{r = 1}^{n} ( a r^{3} + b ) = a \sum_{r = 1}^{n} r^{3} + b n$

  • Coefficients of $r^{3}$* *can come out of the sum
  • Constants are multiplied by $n$

Exam Hint This formula is given in the Formulae Booklet.

Worked Example (a) Find $1^{3}+2^{3}+3^{3}+...+24^{3}$

This is the sum of the first 24 cube numbers Substitute $n=24$ into the formula $\sum_{r = 1}^{n} r^{3} = 1 \frac{1}{4} n^{2} ( n + 1 )^{2}$* *

$\begin{gathered} \sum_{r = 1}^{24} r^{3} & = 1 \frac{1}{4} \times 24^{2} \times ( 24 + 1 )^{2} \\ = 1 \frac{1}{4} \times 24^{2} \times 25^{2} \end{gathered}$

Final answer: 90000

(b) Find and simplify a formula for the sum of the first $n$ even cube numbers.

The sum of even cube numbers is 2³+ 4³+ 6³ + ... The first 5 would end on 10³ The first $n$ would end on $( 2 n )^{3}$

$2^{3} + 4^{3} + 6^{3} +... + ( 2 n )^{3}$* *

The standard formula is for $\sum_{r = 1}^{n} r^{3}$* * To get just the even numbers, replace $r$ with $2r$

$\sum_{r = 1}^{n} ( 2 r )^{3}$

Cube the expression and use $\sum_{r = 1}^{n} a r^{3} = a \sum_{r = 1}^{n} r^{3}$

$\begin{gathered} \sum_{r = 1}^{n} ( 2 r )^{3} & = \sum_{r = 1}^{n} 8 r^{3} \\ = 8 \sum_{r = 1}^{n} r^{3} \end{gathered}$

Now substitute in the standard formula $\sum_{r = 1}^{n} r^{3} = 1 \frac{1}{4} n^{2} ( n + 1 )^{2}$ and simplify

$\begin{gathered} \sum_{r = 1}^{n} ( 2 r )^{3} & = 8 ( 1 \frac{1}{4} n^{2} ( n + 1 )^{2} ) \\ = 2 n^{2} ( n + 1 )^{2} \end{gathered}$

$\begin{gathered} 2 n^{2} ( n + 1 )^{2} \end{gathered}$