K4Maths

📚 Introduction

Building on your knowledge of series and sums involving squares, we now turn to cubed series. These series arise naturally in advanced mathematics, physics, and engineering, and feature prominently in exam questions.

The cubed series involves sums of the form $\sum_{r=1}^n r^3$, as well as more complicated expressions involving cubic terms. Understanding how to break apart and evaluate these sums is a key skill for tackling higher-level problems.

🎯 Learning Objectives

By the end of this lesson, you will be able to:

Understand and use the formula for the sum of cubes of natural numbers
Manipulate and simplify complex cubed series using standard summation techniques
Solve advanced problems involving cubic series with changing limits and algebraic expressions

🔑 Key Concepts

What is a Cubed Series?

A cubed series is a series where each term is raised to the power of three, i.e., it sums $r^3$ for $r$ from $1$ to $n$:

\sum_{r=1}^{n} r^3 = 1^3 + 2^3 + 3^3 + \quad ... \quad + n^3

Unlike sums of natural numbers or squares, cubed series grow rapidly, and their closed-form formula is especially elegant.

📖 The Golden Formula

Sum of Cubes Formula: $$\sum_{r=1}^{n} r^3 = \boxed{\left[\frac{n(n+1)}{2}\right]^2} \quad \text{ Or } \quad \boxed{\frac{1}{4}n^2(n+1)^2}$$
This means the sum of the cubes of the first $n$ natural numbers equals the square of the sum of the first $n$ natural numbers.

\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$$ $$\text{For example: } \sum_{r=1}^4 r^3 = (1^3 + 2^3 + 3^3 + 4^3) = 1 + 8 + 27 + 64 = \boxed{100}$$ $$\left(\frac{4 \times 5}{2}\right)^2 = (10)^2 = \boxed{100}

Cubed Series with Variable Limits

When the lower limit is not $1$, we use subtraction to find the sum: $$\sum_{r=a}^{b} r^3 = \sum_{r=1}^{b} r^3 - \sum_{r=1}^{a-1} r^3$$ This technique is essential for problems where the series does not start at $1$.

💡 Remember this:

For any series of the form $\sum_{r=1}^{n} (\text{expression})$, always look to break it into standard sums using linearity: split into sums of $r^3$, $r^2$, $r$, and constants.

Standard Summation Formulas

Here are the key formulas you should know as of now, and if you dont, make sure to memorise them:

\sum_{r=1}^n r = \frac{n(n+1)}{2}$$ $$\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{r=1}^n r^3 = \left[\frac{n(n+1)}{2}\right]^2$$ $$\sum_{r=1}^n c = cn

With these, you can tackle almost any algebraic series. I highly reccomend having prior knowledge in series before persuing this lesson, you can learn basics and squares of series here.

📝 Worked Examples

📋 Example 1: Simple Cubed Series

Question: Find the value of $$\sum_{r=1}^{8} r^3$$

Solution:

Step 1:

Recall and apply the formula:

\sum_{r=1}^{8} r^3 = \left(\frac{n(n+1)}{2}\right)^2 \rightarrow \left(\frac{8 \times 9}{2}\right)^2 = (36)^2 = 1296

Final Answer:

So the sum is:

\boxed{1296}

📋 Example 2: Cubed Series with Non-Unit Lower Limit

Question: Calculate $$\sum_{r=5}^{10} r^3$$

Solution:

Step 1:

Express as difference of two sums:

\sum_{r=5}^{10} r^3 = \sum_{r=1}^{10} r^3 - \sum_{r=1}^{4} r^3

Step 2:

Evaluate each sum using the formula:

\sum_{r=1}^{10} r^3 = \left(\frac{10 \times 11}{2}\right)^2 = (55)^2 = 3025$$ $$\sum_{r=1}^{4} r^3 = (10)^2 = 100

Final Answer:

Subtract to get:

3025 - 100 = \boxed{2925}

📋 Example 3: Complex Algebraic Cubed Series

Question: Let: $$S(n) = \sum_{r=1}^n r(r+3)(2r-1)$$ Show that $$S(n) = \frac{1}{6} n(n+1)(3n^2 + an + b) \text{ for some integers a and b}$$ Then, find $$a \text{ and } b$$

Solution:

Step 1:

Expand the brackets inside the summation:

$$r(r+3)(2r-1) = r(2r^2 + 6r - r - 3)$$ $$ = r(2r^2 + 5r - 3)$$ $$= 2r^3 + 5r^2 - 3r$$

Step 2:

Now it turns into this:

S(n) = \sum_{r=1}^n [2r^3 + 5r^2 - 3r]$$ $$\text{Split it into different sums: } 2\sum_{r=1}^n r^3 + 5\sum_{r=1}^n r^2 - 3\sum_{r=1}^n r

Remember we can take the coefficient out of the summation/series

Step 3:

Apply the formulas:

\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2 \leftarrow \text{cubic}$$ $$\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6} \leftarrow \text{square}$$ $$\sum_{r=1}^n r = \frac{n(n+1)}{2} \leftarrow \text{constant}

Step 4:

Substitute the formulas and simplify:

\text{Cubic} \rightarrow 2\left(\frac{n(n+1)}{2}\right)^2$$ $$\text{Square} \rightarrow 5\left[\frac{n(n+1)(2n+1)}{6}\right]$$ $$\text{Constant} \rightarrow -3\left[\frac{n(n+1)}{2}\right]$$ $$\text{We now have}$$ $$S(n) = 2\left(\frac{n(n+1)}{2}\right)^2 + 5\left[\frac{n(n+1)(2n+1)}{6}\right] - 3\left[\frac{n(n+1)}{2}\right]$$ $$\text{From:}$$ $$S(n) = \sum_{r=1}^n [2r^3 + 5r^2 - 3r] = 2\sum_{r=1}^n r^3 + 5\sum_{r=1}^n r^2 - 3\sum_{r=1}^n r

Step 5:

Put the coefficient in the numerator then set all terms to a common denominator ($6$) to combine:

\text{Coefficients in the numerator:}$$ $$= \frac{3n^2(n+1)^2}{6} + \frac{5n(n+1)(2n+1)}{6} - \frac{9n(n+1)}{6}$$ $$\text{Now find common denominator by multiplying the 1st and 3rd}$$ $$= \frac{9n^2(n+1)^2}{6} + \frac{5n(n+1)(2n+1)}{6} - \frac{9n(n+1)}{6}$$ $$\text{Combine to one:}$$ $$= \frac{[9n^2(n+1)^2 + 5n(n+1)(2n+1) - 9n(n+1)]}{6} $$ $$\equiv \frac{1}{6} [9n^2(n+1)^2 + 5n(n+1)(2n+1) - 9n(n+1)]

$\equiv$ means 'equivalent to'.

Step 6:

Simplify by factoring $n(n+1)$ and adding like-terms:

\text{We have: } \frac{1}{6} [9n^2(n+1)^2 + 5n(n+1)(2n+1) - 9n(n+1)]$$ $$\text{Factoring } n(n+1) \text{ out: }$$ $$\frac{1}{6}n(n+1)[3n(n+1) + 5(2n+1) - 9]$$ $$\text{Distributing the coefficients next to the brackets: }$$ $$= \frac{1}{6}n(n+1)[3n^2 + 3n + 10n + 5 - 9]$$ $$\text{Combining like-terms inside the brackets:}$$ $$= \frac{1}{6}n(n+1)[3n^2 + 13n - 4]

Final Answer:

So, we have:

\boxed{S(n) = \frac{1}{6} n(n+1)(3n^2 + 13n - 4) \quad \therefore a = 13, b = -4.}

$\therefore$ means 'therefore'.

📋 Example 4: Applying Limits to Advanced Cubic Series

Question: Hence compute $\sum_{r=11}^{40} r(r+3)(2r-1)$.

Solution:

Step 1:

Acknowledge limits are starting at $11$:

\sum_{\boxed{r=11}}^{40} r(r+3)(2r-1) =$$ $$= \sum_{r=1}^{40} r(r+3)(2r-1) - \sum_{r=1}^{10} r(r+3)(2r-1)

Step 2:

Expand brackets and break it up into multiple summations:

\text{Expanding:}$$ $$r(r+3)(2r-1) = 2r^3 + 7r^2 + 3r$$ $$\text{So we have:}$$ $$\sum_{r=1}^{40} (2r^3 + 7r^2 + 3r) - \sum_{r=1}^{10} (2r^3 + 7r^2 + 3r)$$ $$\text{Splitting:}$$ $$\underbrace{2\sum_{r=1}^{40} r^3 + 7\sum_{r=1}^{40} r^2 + 3\sum_{r=1}^{40} r}_{\text{let this be equation 1}}\quad \underbrace{- \left(2\sum_{r=1}^{10} r^3 + 7\sum_{r=1}^{10} r^2 + 3\sum_{r=1}^{10} r\right)}_{\text{let this be equation 2}}

Looks daunting, but its quite simple as long as you practice.

Step 3:

For equation 1:

\text{First series: } 2\sum_{r=1}^{40} r^3 = 2\left(\frac{40 \times 41}{2}\right)^2 = 2(820)^2 = \underline{1342400}$$ $$\text{Second series: } 7\sum_{r=1}^{40} r^2 = 7\left[\frac{40 \times 41 \times 81}{6}\right] = 7(22140) = \underline{154980}$$ $$\text{Third series: } 3\sum_{r=1}^{40} r = 3\left[\frac{40 \times 41}{2}\right] = 3(820) = \underline{2460}$$ $$\text{So equation 1 is: } \underline{1342400 + 154980 + 2460 = 1499840}

Step 4:

Equation 2 now: - \left(2\sum_{r=1}^{10} r^3 + 7\sum_{r=1}^{10} r^2 + 3\sum_{r=1}^{10} r\right

\text{First series: } 2\sum_{r=1}^{10} r^3 = 2\left(\frac{10 \times 11}{2}\right)^2 = 2(55)^2 = \underline{6050}$$ $$\text{Second series: } 7\sum_{r=1}^{10} r^2 = 7\left[\frac{10 \times 11 \times 21}{6}\right] = 7(385) = \underline{2695}$$ $$\text{Third series: } 3\sum_{r=1}^{10} r = 3\left[\frac{10 \times 11}{2}\right] = 3(55) = \underline{165}$$ $$\text{So equation 2 is: } \underline{-(6050 + 2695 + 165) = -8910}

Final Answer:

Now combine both equations:

\boxed{\text{Final Answer: } 1499840 - 8910 = \underline{1490930}}

⚠️ Common Mistakes

🚫 Mistake 1: Not splitting algebraic expressions into standard summations

What students often do wrong: Students may attempt to sum a complicated cubic directly rather than expand and break it into sums of $r^3$, $r^2$, and $r$.

Why it's wrong: This prevents you from using powerful formulas and leads to errors.

How to avoid it: Always expand the cubic, split into standard sums, and apply the formulas.

🚫 Mistake 2: Forgetting to adjust the lower limit

When the series starts at a number other than 1, use subtraction to find the correct sum.

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Question 1

(A) Calculate $\sum_{r=1}^{12} r^3$
(B) Calculate $\sum_{r=7}^{15} r^3$

🔍 Show Solution

(A) $\sum_{r=1}^{12} r^3 = \left(\frac{12 \times 13}{2}\right)^2 = (78)^2 = \boxed{6084}$
(B) $\sum_{r=7}^{15} r^3 = \sum_{r=1}^{15} r^3 - \sum_{r=1}^{6} r^3 = (120)^2 - (21)^2 = 14400 - 441 = \boxed{13959}$

🎯 Practice Question 2

Let $T(n) = \sum_{r=1}^n (4r^3 - 2r^2 + 7)$.
(A) Express $T(n)$ in terms of standard summations.
(B) Find $T(20)$.

🔍 Show Solution

(A) $T(n) = 4\sum_{r=1}^n r^3 - 2\sum_{r=1}^n r^2 + 7n$
(B) For $n=20$:
$\sum_{r=1}^{20} r^3 = (210)^2 = 44100$
$\sum_{r=1}^{20} r^2 = \frac{20 \times 21 \times 41}{6} = 2870$
So,
$T(20) = 4 \times 44100 - 2 \times 2870 + 7 \times 20 = 176400 - 5740 + 140 = \boxed{170800}$

🎯 Practice Question 3 (Advanced)

Let $Q(n) = \sum_{r=1}^n r(r-2)(r+4)$.
(A) Express $Q(n)$ as a sum of cubes, squares, and $r$.
(B) Find a closed-form for $Q(n)$.

🔍 Show Solution

(A) Expand:
$r(r-2)(r+4) = r(r^2 + 4r - 2r - 8) = r(r^2 + 2r - 8) = r^3 + 2r^2 - 8r$
$Q(n) = \sum_{r=1}^n (r^3 + 2r^2 - 8r) = \sum_{r=1}^n r^3 + 2\sum_{r=1}^n r^2 - 8\sum_{r=1}^n r$
(B) Substitute formulas:
$Q(n) = \left(\frac{n(n+1)}{2}\right)^2 + 2\left[\frac{n(n+1)(2n+1)}{6}\right] - 8\left[\frac{n(n+1)}{2}\right]$
$= \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} - 4n(n+1)$

🎯 Challenge question:

Let $$\sum_{k=2}^{n} \bigl(k^3 - 6k^2 + 11k - 6\bigr) \;=\; 120.$$ Given that one of the roots of the resulting cubic equation is $n=3$, find the value(s) of $n$.

🔍 Show Solution

\section*{Step-by-Step Solution} $$\text{Step 1: Write the sum in terms of known formulas}$$ $$\text{Set: } S = \sum_{k=2}^{n} (k^3 - 6k^2 + 11k - 6) = \sum_{k=2}^{n} k^3 - 6\sum_{k=2}^{n} k^2 + 11\sum_{k=2}^{n} k - 6\sum_{k=2}^{n}1$$ $$\text{Use the identity } \sum_{k=2}^n f(k)=\sum_{k=1}^n f(k)-f(1) \text{ with the standard formulas:}$$ $$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n}k^3 = \left(\frac{n(n+1)}{2}\right)^2$$ $$\text{Thus:}$$ $$\sum_{k=2}^{n}k^3 = \left(\frac{n(n+1)}{2}\right)^2 - 1$$ $$\sum_{k=2}^{n}k^2 = \frac{n(n+1)(2n+1)}{6} - 1$$ $$\sum_{k=2}^{n}k = \frac{n(n+1)}{2} - 1$$ $$\sum_{k=2}^{n}1 = n - 1$$ $$\text{Step 2: Simplify term by term}$$ $$\text{Substitute back:}$$ $$S = \left(\frac{n(n+1)}{2}\right)^2 - 1 - 6\left(\frac{n(n+1)(2n+1)}{6}-1\right) + 11\left(\frac{n(n+1)}{2}-1\right) - 6(n-1)$$ $$\text{Simplifying each term:}$$ $$= \frac{n^2(n+1)^2}{4} - n(n+1)(2n+1) + \frac{11n(n+1)}{2} - 6n$$ $$\text{Write everything over a common denominator (4):}$$ $$S(n) = \frac{1}{4}(n^4 - 6n^3 + 11n^2 - 6n)$$ $$\text{Step 3: Set the sum equal to 120}$$ $$\frac{1}{4}(n^4 - 6n^3 + 11n^2 - 6n) = 120$$ $$n^4 - 6n^3 + 11n^2 - 6n = 480$$ $$n^4 - 6n^3 + 11n^2 - 6n - 480 = 0$$ $$\text{Step 4: Factor the polynomial}$$ $$\text{Observe that: } n^4 - 6n^3 + 11n^2 - 6n = n(n^3 - 6n^2 + 11n - 6) = n(n-1)(n-2)(n-3)$$ $$\text{Thus the equation becomes: } n(n-1)(n-2)(n-3) = 480$$ $$\text{Step 5: Solve for integer } n$$ $$\text{Test small integer values:}$$ $$n=5 \Rightarrow 5 \cdot 4 \cdot 3 \cdot 2 = 120 \text{ (too small)}$$ $$n=6 \Rightarrow 6 \cdot 5 \cdot 4 \cdot 3 = 360 \text{ (too small)}$$ $$n=7 \Rightarrow 7 \cdot 6 \cdot 5 \cdot 4 = 840 \text{ (too large)}$$ $$\text{For } n=8: 8 \cdot 7 \cdot 6 \cdot 5 = 1680 \text{ (too large)}$$ $$\text{Since we need } n(n-1)(n-2)(n-3) = 480 \text{ we can check } n = 6.4 \text{ approximately}$$ $$\boxed{\text{Final Answer: } n = 8}$$

📋 Summary

🎯 Key Takeaways

The sum of cubes is given by $\sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2$
Complex cubic series can be broken down into sums of cubes, squares, and linear terms
Mastery of series techniques lets you solve advanced algebraic summations

📚 What's Next?

Practice more cubic and higher power series, and explore applications in combinatorics and calculus. Next, consider learning about quartic series and their uses in mathematical proofs and problems.