Adding up the Terms of a Sequence Revision Notes for HSC SSCE Mathematics Advanced

Adding up the Terms of a Sequence

Introduction: why we add sequence terms

When working with sequences, we frequently need to add their terms together. This is a crucial skill with many practical applications in mathematics and real-world scenarios.

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Real-world example: Consider a boulder falling from a cliff. It falls $5$ metres in the first second, $15$ metres in the second second, $25$ metres in the third second, and so on. To find the total distance fallen in the first $10$ seconds, we need to add up these distances:

$5 + 15 + 25 + 35 + \ldots + 95 = 500 \text{ metres}$

This demonstrates why learning to sum sequence terms efficiently is so important.

Understanding partial sums and their notation

What is $S_n$ ?

For any sequence $T_1, T_2, T_3, \ldots$ , we use the symbol $S_n$ to represent the sum of the first $n$ terms. This is called the partial sum because it represents part of the sequence.

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Key formula:

$S_n = T_1 + T_2 + T_3 + \ldots + T_n$

This notation is fundamental to working with sequence sums throughout mathematics.

Different names for $S_n$

The sum $S_n$ can be referred to in several ways:

The sum of the first $n$ terms of the sequence
The sum to $n$ terms of the sequence
The $n$ th partial sum of the sequence (where 'partial' means we're only considering part of the sequence)

Example with the falling boulder

Using our boulder sequence $5, 15, 25, 35, \ldots$ , the sum of the first $10$ terms is:

$S_{10} = 5 + 15 + 25 + 35 + 45 + 55 + 65 + 75 + 85 + 95 = 500$

This is the $10$ th partial sum of the sequence.

The sequence of partial sums

An interesting feature is that the partial sums themselves form a new sequence. Let's see how this works.

Building the partial sums sequence

Starting with the sequence $5, 15, 25, 35, \ldots$ , we can calculate successive partial sums:

$S_1 = 5$

$S_2 = 5 + 15 = 20$

$S_3 = 5 + 15 + 25 = 45$

$S_4 = 5 + 15 + 25 + 35 = 80$

Notice how each partial sum includes all the previous terms plus one new term. The sequence of partial sums is $5, 20, 45, 80, \ldots$

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Worked Example: Completing a Partial Sums Table

Question: Copy and complete this table for the successive sums of a sequence.

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$T_n$	$5$	$15$	$25$	$35$	$45$	$55$	$65$	$75$	$85$	$95$
$S_n$

Solution: Each entry for $S_n$ represents the sum of all the terms $T_n$ up to that position.

Working:

$S_1 = 5$
$S_2 = 5 + 15 = 20$
$S_3 = 5 + 15 + 25 = 45$
$S_4 = 5 + 15 + 25 + 35 = 80$
And so on...

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Exam tip: When calculating partial sums in a table, always start from the left and add one new term at a time. This reduces calculation errors.

Recovering the original sequence from partial sums

Sometimes we know the partial sums $S_n$ and need to find the original sequence $T_n$ . We can do this by taking successive differences.

The method

This technique allows us to work backwards from partial sums to recover the original sequence terms.

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First term: $T_1 = S_1$

Subsequent terms: $T_n = S_n - S_{n-1}$ for $n \geq 2$

This makes sense because $S_n$ contains all terms up to $T_n$ , and $S_{n-1}$ contains all terms up to $T_{n-1}$ . Their difference must be $T_n$ .

Why this works

Think about what the partial sums represent:

$S_n = T_1 + T_2 + T_3 + \ldots + T_n$
$S_{n-1} = T_1 + T_2 + T_3 + \ldots + T_{n-1}$

When we subtract: $S_n - S_{n-1} = T_n$

All the earlier terms cancel out, leaving only $T_n$ .

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Worked Example: Finding Terms from Partial Sums

Question: By taking successive differences, list the terms of the original sequence.

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$T_n$
$S_n$	$1$	$5$	$12$	$22$	$35$	$51$	$70$	$92$	$117$	$145$

Solution: Each entry for $T_n$ is found by taking the difference between two consecutive values of $S_n$ .

Working:

$T_1 = S_1 = 1$
$T_2 = S_2 - S_1 = 5 - 1 = 4$
$T_3 = S_3 - S_2 = 12 - 5 = 7$
$T_4 = S_4 - S_3 = 22 - 12 = 10$

The pattern continues, giving us the sequence $1, 4, 7, 10, 13, 16, 19, 22, 25, 28$ . This is an arithmetic sequence with first term $1$ and common difference $3$ .

Algebraic proof example

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Worked Example: Algebraic Proof

Question: Prove algebraically that if the partial sums $S_n$ of a sequence are the successive squares (that is, $S_n = n^2$ ), then the sequence $T_n$ is the sequence of odd numbers.

Solution:

We are given that $S_n = n^2$ .

For the first term:

$T_1 = S_1 = 1^2 = 1$

This is indeed the first odd number.

For $n \geq 2$ :

T_n = S_n - S_{n-1} = n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = n^2 - n^2 + 2n - 1 = 2n - 1

This is the general formula for the $n$ th odd number, confirming our result.

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Exam tip: When expanding $(n-1)^2$ , remember it equals $n^2 - 2n + 1$ , not $n^2 - 1$ . This is a common mistake.

Sigma notation

Sigma notation provides a concise way to write sums. The symbol $\sum$ (the Greek capital letter sigma) stands for 'sum'.

Understanding the notation

The general form is:

$\sum_{n=a}^{b} T_n = T_a + T_{a+1} + T_{a+2} + \ldots + T_b$

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Reading this notation:

The $n = a$ at the bottom tells us the starting value
The $b$ at the top tells us the ending value
$T_n$ is the general term we're summing
We evaluate $T_n$ for each integer from $a$ to $b$ , then add up all the results

Examples of sigma notation

Example 1:

$\sum_{n=2}^{5} n^2 = 2^2 + 3^2 + 4^2 + 5^2$

$= 4 + 9 + 16 + 25$

$= 54$

This sum has $4$ terms (from $n = 2$ to $n = 5$ ).

Example 2:

$\sum_{n=6}^{10} n^2 = 6^2 + 7^2 + 8^2 + 9^2 + 10^2$

$= 36 + 49 + 64 + 81 + 100$

$= 330$

General form for sequences

For a sequence $T_1, T_2, T_3, \ldots$ :

$\sum_{n=5}^{20} T_n = T_5 + T_6 + T_7 + T_8 + \ldots + T_{20}$

Any two integers can be used as the lower and upper limits.

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Worked Example: Evaluating Sums

Question: Evaluate these sums.

a) $\displaystyle\sum_{n=4}^{7} (5n + 1)$

b) $\displaystyle\sum_{n=1}^{5} (-2)^n$

Solution:

a) First, substitute each value of $n$ from $4$ to $7$ into the expression $5n + 1$ :

$\sum_{n=4}^{7} (5n + 1) = (5 \times 4 + 1) + (5 \times 5 + 1) + (5 \times 6 + 1) + (5 \times 7 + 1)$

$= 21 + 26 + 31 + 36$

$= 114$

b) Substitute each value of $n$ from $1$ to $5$ into $(-2)^n$ :

$\sum_{n=1}^{5} (-2)^n = (-2)^1 + (-2)^2 + (-2)^3 + (-2)^4 + (-2)^5$

$= -2 + 4 - 8 + 16 - 32$

$= -22$

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Exam tip: When dealing with negative bases and powers, be careful with signs. Remember that $(-2)^2 = 4$ (positive) but $(-2)^3 = -8$ (negative).

Understanding series

The term series is related to sequences but has a specific meaning in mathematics.

What is a series?

A series refers to the process of adding up terms of a sequence. More precisely, a series is the sequence of partial sums.

When we write "the series $1 + 4 + 9 + \ldots$ ", we mean we're considering the successive partial sums:

$S_1 = 1$

$S_2 = 1 + 4 = 5$

$S_3 = 1 + 4 + 9 = 14$

$S_4 = 1 + 4 + 9 + 16 = 30$

And so on.

Formal definition

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Given a sequence $T_1, T_2, T_3, \ldots$ , the corresponding series is the sequence of partial sums:

$S_1 = T_1$

$S_2 = T_1 + T_2$

$S_3 = T_1 + T_2 + T_3$

$S_4 = T_1 + T_2 + T_3 + T_4$

And so on.

The series is therefore another sequence: $S_1, S_2, S_3, S_4, \ldots$

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Key Points to Remember:

$S_n$ represents the sum of the first $n$ terms of a sequence: $S_n = T_1 + T_2 + \ldots + T_n$
To find partial sums in a table, add each new term to the previous sum as you move along
To recover the original sequence from partial sums, use $T_1 = S_1$ and $T_n = S_n - S_{n-1}$ for $n \geq 2$
Sigma notation ( $\sum$ ) provides a concise way to write sums, with the lower limit showing where to start and the upper limit showing where to end
A series is the sequence of partial sums formed by adding up terms of a sequence

Adding up the Terms of a Sequence (HSC SSCE Mathematics Advanced): Revision Notes