Sums of Integers, Squares & Cubes (Edexcel A-Level Further Mathematics): Revision Notes

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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:

  1. Sum of the first nn integers.
  2. Sum of the squares of the first nn integers.
  3. Sum of the cubes of the first nn 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

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

Sum of Squares

∑r=1nr2=12+22+32+⋯+n2=n(n+1)(2n+1)6\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

∑r=1nr3=13+23+33+⋯+n3=[n(n+1)2]2\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 ∑r=110r2\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 ∑r=1nr2\sum_{r=1}^n r^2

∑r=1nr2=n(n+1)(2n+1)6.\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}.

Step 2: Substitute n=10n = 10

∑r=110r2=10(10+1)(2×10+1)6\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \times 10 + 1)}{6}

Step 3**: Simplify**

.∑r=110r2=10×11×216=23106=385.\sum_{r=1}^{10} r^2 = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385

Final Answer:

∑r=110r2=:highlight[385].\sum_{r=1}^{10} r^2 = :highlight[385].
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Example 2

Evaluate ∑r=18(r+r2)\sum_{r=1}^8 (r + r^2)

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

∑r=18(r+r2)=∑r=18r+∑r=18r2\sum_{r=1}^8 (r + r^2) = \sum_{r=1}^8 r + \sum_{r=1}^8 r^2

Step 2**: Use the formulas for** ∑r\sum r and ∑r2\sum r^2

For ∑r=1nr\sum_{r=1}^n r:

∑r=18r=n(n+1)2=8(8+1)2\sum_{r=1}^8 r = \frac{n(n+1)}{2} = \frac{8(8+1)}{2} =8×92=36= \frac{8 \times 9}{2} = 36

For ∑r=1nr2\sum_{r=1}^n r^2:

∑r=18r2=n(n+1)(2n+1)6=8(8+1)(2×8+1)6\sum_{r=1}^8 r^2 = \frac{n(n+1)(2n+1)}{6} = \frac{8(8+1)(2 \times 8 + 1)}{6}

Simplify:

∑r=18r2=8×9×176=12246=204\sum_{r=1}^8 r^2 = \frac{8 \times 9 \times 17}{6} = \frac{1224}{6} = 204

Step 3**: Add the results**

∑r=18(r+r2)=36+204=240\sum_{r=1}^8 (r + r^2) = 36 + 204 = 240

Final Answer:

∑r=18(r+r2)=:highlight[240]\sum_{r=1}^8 (r + r^2) = :highlight[240]
infoNote

Example 3

Evaluate ∑r=15(2r3+3r)\sum_{r=1}^{5} (2r^3 + 3r)

Step 1**: Split the summation**

∑r=15(2r3+3r)=2∑r=15r3+3∑r=15r\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 ∑r=1nr3\sum_{r=1}^n r^3

∑r=15r3=[n(n+1)2]2=[5(5+1)2]2\sum_{r=1}^5 r^3 = \left[\frac{n(n+1)}{2}\right]^2 = \left[\frac{5(5+1)}{2}\right]^2 =[5×62]2=152=225= \left[\frac{5 \times 6}{2}\right]^2 = 15^2 = 225

For ∑r=1nr\sum_{r=1}^n r

∑r=15r=n(n+1)2=5(5+1)2\sum_{r=1}^5 r = \frac{n(n+1)}{2} = \frac{5(5+1)}{2} =5×62=15= \frac{5 \times 6}{2} = 15

Step 3**: Compute the weighted sums**

∑r=15r3=2×225=450\sum_{r=1}^{5} r^3 = 2 \times 225 = 4503∑r=15r=3×15=453\sum_{r=1}^{5} r = 3 \times 15 = 45

Step 4**: Add the results**

∑r=15(2r3+3r)=450+45=495\sum_{r=1}^{5} (2r^3 + 3r) = 450 + 45 = 495

Final Answer:

∑r=15(2r3+3r)=:highlight[495]\sum_{r=1}^{5} (2r^3 + 3r) = :highlight[495]

Note Summary

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Common Mistakes

  1. Forgetting to substitute correctly: Errors occur when substituting nn into the formulas for sums.
  2. Misapplying the formulas: Confusing the formulas for ∑r2\sum r^2 and ∑r3\sum r^3
  3. Not simplifying properly: Mistakes often arise during algebraic simplifications, particularly when dealing with large numbers.
  4. Ignoring coefficients in weighted sums: Failing to multiply by constants (e.g., the 22 in 2r3^3).
  5. Combining terms incorrectly: When breaking a complex series into parts, errors occur when recombining the results.
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Key Formulas

  1. The sum of integers:
∑r=1nr=n(n+1)2\sum_{r=1}^n r = \frac{n(n+1)}{2}
  1. The sum of squares:
∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}
  1. The sum of cubes:
∑r=1nr3=[n(n+1)2]2\sum_{r=1}^n r^3 = \left[\frac{n(n+1)}{2}\right]^2
  1. Weighted sum:
∑r=1n(ar2+br)=a∑r=1nr2+b∑r=1nr\sum_{r=1}^n (ar^2 + br) = a\sum_{r=1}^n r^2 + b\sum_{r=1}^n r
  1. Simplifying summation notation: Break complex series into simpler components and sum each part separately.

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