3.2.1 Sums of Integers, Squares & Cubes

Introduction

Summing integers, their squares, and their cubes is a fundamental topic. These sums are represented using summation notation and are essential for solving problems in series and sequences.

The general forms of the sums are:

Sum of the first $n$ integers.
Sum of the squares of the first $n$ integers.
Sum of the cubes of the first $n$ integers. In this section, we'll derive these formulas, explore their applications, and use them to evaluate more complex series.

Key Formulas for Sums

Sum of Integers

\sum_{r=1}^n r = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}

Sum of Squares

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

Sum of Cubes

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

These formulas provide the foundation for summing more complex series by breaking them into parts or simplifying their structure.

Worked Examples

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Example 1

Compute $\sum_{r=1}^{10} r^2$

We are tasked with finding the sum of squares of the first 10 integers.

Step 1: Write the formula for $\sum_{r=1}^n r^2$

\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}.

Step 2: Substitute $n = 10$

\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \times 10 + 1)}{6}

Step 3**: Simplify**

.\sum_{r=1}^{10} r^2 = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385

Final Answer:

\sum_{r=1}^{10} r^2 = :highlight[385].

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Example 2

Evaluate $\sum_{r=1}^8 (r + r^2)$

Step 1**: Break the summation into separate parts**

\sum_{r=1}^8 (r + r^2) = \sum_{r=1}^8 r + \sum_{r=1}^8 r^2

Step 2**: Use the formulas for** $\sum r$ and $\sum r^2$

For $\sum_{r=1}^n r$ :

\sum_{r=1}^8 r = \frac{n(n+1)}{2} = \frac{8(8+1)}{2}

= \frac{8 \times 9}{2} = 36

For $\sum_{r=1}^n r^2$ :

\sum_{r=1}^8 r^2 = \frac{n(n+1)(2n+1)}{6} = \frac{8(8+1)(2 \times 8 + 1)}{6}

Simplify:

\sum_{r=1}^8 r^2 = \frac{8 \times 9 \times 17}{6} = \frac{1224}{6} = 204

Step 3**: Add the results**

\sum_{r=1}^8 (r + r^2) = 36 + 204 = 240

Final Answer:

\sum_{r=1}^8 (r + r^2) = :highlight[240]

infoNote

Example 3

Evaluate $\sum_{r=1}^{5} (2r^3 + 3r)$

Step 1**: Split the summation**

\sum_{r=1}^{5} (2r^3 + 3r) = 2\sum_{r=1}^{5} r^3 + 3\sum_{r=1}^{5} r

Step 2**: Use the formulas**

For $\sum_{r=1}^n r^3$

\sum_{r=1}^5 r^3 = \left[\frac{n(n+1)}{2}\right]^2 = \left[\frac{5(5+1)}{2}\right]^2

= \left[\frac{5 \times 6}{2}\right]^2 = 15^2 = 225

For $\sum_{r=1}^n r$

\sum_{r=1}^5 r = \frac{n(n+1)}{2} = \frac{5(5+1)}{2}

= \frac{5 \times 6}{2} = 15

Step 3**: Compute the weighted sums**

\sum_{r=1}^{5} r^3 = 2 \times 225 = 450

3\sum_{r=1}^{5} r = 3 \times 15 = 45

Step 4**: Add the results**

\sum_{r=1}^{5} (2r^3 + 3r) = 450 + 45 = 495

Final Answer:

\sum_{r=1}^{5} (2r^3 + 3r) = :highlight[495]

Note Summary

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Common Mistakes

Forgetting to substitute correctly: Errors occur when substituting $n$ into the formulas for sums.
Misapplying the formulas: Confusing the formulas for $\sum r^2$ and $\sum r^3$
Not simplifying properly: Mistakes often arise during algebraic simplifications, particularly when dealing with large numbers.
Ignoring coefficients in weighted sums: Failing to multiply by constants (e.g., the $2$ in 2r $^3$ ).
Combining terms incorrectly: When breaking a complex series into parts, errors occur when recombining the results.

infoNote

Key Formulas

The sum of integers:

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

The sum of squares:

\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}

The sum of cubes:

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

Weighted sum:

\sum_{r=1}^n (ar^2 + br) = a\sum_{r=1}^n r^2 + b\sum_{r=1}^n r

Simplifying summation notation: Break complex series into simpler components and sum each part separately.

Sums of Integers, Squares & Cubes Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

3.2.1 Sums of Integers, Squares & Cubes

Introduction

The general forms of the sums are:

Key Formulas for Sums

Sum of Integers

Sum of Squares

Sum of Cubes

Worked Examples

Example 1

Example 2

Example 3

Note Summary

Common Mistakes

Key Formulas

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