Square Series

Understanding the sum of a square series.

📚 Further Maths 🎯 Difficulty: ⭐⭐ ⏱️ Reading time: 30 minutes 📋 Edexcel & AQA

📚 Introduction

In this lesson, we will be looking at Square series, which look like this: $\sum^n_{r=1} r^2$. We'll also understand it and know how to solve problems for it. You must understand the previous lesson (series basics) before this one.

🎯 Learning Objectives

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

Understand the concept of square series
Apply the formula to solve problems involving square series

🔑 Key Concepts

What is the Square series?

A square series is a natural number series that involves the squares of the first n natural numbers. It is represented as:$$\sum^n_{r=1} r^2$$.
This means we take every natural number from $r$ (which is $1$) to $n$ and square each of them, then add them all together. For example: $$\sum^{4}_{r=1} r^2 \quad \text{means: } \quad 1^2 + 2^2 + 3^2 + 4^2$$ $$\text{Which equals: } 1 + 4 + 9 + 16 = 30$$

📖 The Golden Formula:

Sum of Squares Formula: $$\sum^n_{r=1} r^2 = \frac{n(n + 1)(2n + 1)}{6} \quad\quad \text{Or} \quad\quad \frac{1}{6}\times n(n + 1)(2n + 1)$$
Heres a mini example below, solving the following: $$\sum^{25}_{r=1} r^2$$

\sum^n_{r=1} r^2 = \frac{n(n + 1)(2n + 1)}{6}$$ $$\sum^{25}_{r=1} r^2 = \frac{25(25 + 1)(2\times25 + 1)}{6}$$ $$\frac{25(25 + 1)(2\times25 + 1)}{6} = \frac{25\times26\times51}{6} = 5525$$ $$\text{So, } 1^2 + 2^2 + 3^2 + ... + 25^2 = 5525

Another Concept:

Like the previous lesson, you can also add and subtract mutiple series'. This is especially helpful for complex questions. Heres a simple example: $$\sum^{10}_{r=1} (r^2 + r) = \sum^{10}_{r=1} r^2 + \sum^{10}_{r=1} r$$ $$\text{Notice how the }r^2 \text{ & } r \text{ get its own series? We can solve them both using the known formulas.}$$

💡 Pro Tip

Dont forget to break up your series. Its goated. Trust me. It makes life easier.

Algrbraic Square Series

You will encounter questions with algebraic expressions in the series. For example: $$\sum^{n}_{r=1} (2r^2 + 3)$$ $$\text{We can break this up into:}$$ $$2\sum^{n}_{r=1} r^2 + \sum^{n}_{r=1} 3$$ $$\text{We just took the 2 and put it in front, then we broke up the square series}$$ $$\text{ and added the series with the constant, 3.}$$

💡Reccommendation

I highly reccomend taking my previous lesson on this. If this is your first lesson, its likely you're confused. Click on this sentence to go to that lesson.

📝 Worked Examples

📋 Example 1: Solving Simple Square Series

Question: Solve $$\sum^{52}_{15} [5r^2 + r + 5]$$

Solution:

Step 1:

Break it into different series', noting that it starts at 15.

\sum^{52}_{r=15} [r^2 + r + 5] =5 \sum^{52}_{r=15} r^2 + \sum^{52}_{r=15} r + \sum^{52}_{r=15} 5

Above, we split each term into their own series and took 5 outside for a walk.

Step 2:

Now we can solve each series using known formulas, and take them away from r=1 to 14, as the lower limit starts at 15. Remember the formula for squares: $$\sum^{n}_{r=1} r^2 = \frac{n(n + 1)(2n + 1)}{6}$$

\text{First series:}$$ $$5 \sum^{52}_{r=15} r^2 = 5 \sum^{52}_{r=1} r^2 - 5\sum^{14}_{r=1} r^2 $$ $$ \text{Which: }$$ $$ 5 \left(\frac{52(52 + 1)(2\times52 + 1)}{6}\right) - 5\left(\frac{14(14 + 1)(2\times14 + 1)}{6}\right)$$ $$ = \underline{23470}

Step 3:

Now we can solve the second series. Remember the formula for sums: $$\sum^{n}_{r=1} r = \frac{n(n + 1)}{2}$$

\sum^{52}_{r=15} r = \sum^{52}_{r=1} r - \sum^{14}_{r=1} r$$ $$ \text{Which: }$$ $$ \sum^{52}_{r=1} r - \sum^{14}_{r=1} r = \left(\frac{52(52 + 1)}{2}\right) - \left(\frac{14(14 + 1)}{2}\right)$$ $$ = \underline{1245}

Step 4:

Now we can solve the third series. Remember the formula for constants: $$\sum^{n}_{r=1} a = an$$

\sum^{52}_{r=15} 5 = \sum^{52}_{r=1} 5 - \sum^{14}_{r=1} 5$$ $$ \text{Which: }$$ $$ 5(52) - 5(14) = \underline{190}

Final Step & Answer:

Now we can add all the answers together to get the final answer.

\boxed{ \underbrace{23470}_{\text{First series}} + \underbrace{ 1245}_{\text{Second series}} + \underbrace{ 190}_{\text{Third series}} \;= \underbrace{24905}_{\text{Final answer}}}

📋 Example 2: Algebraic Series:

$$\underline{ \text{(A) Show } \sum^n_{r=1} [r^2+r-1] = \frac{n(n^2+3n-1)}{3} \quad n, r \in \mathbb{N} }$$ $$\text{(B) Hence, verify this result with: } n = 3$$

Solution:

Part A - Step 1:

For part A, lets break it up into different series' using our known formulas:

\sum^{n}_{r=1} [r^2 + r - 1] = \sum^{n}_{r=1} r^2 + \sum^{n}_{r=1} r - \sum^{n}_{r=1} 1

Above, we split each term into their own series.

Part A - Step 2:

Now we can solve each series using known formulas. Remember the formula for squares:

\text{First series:}$$ $$\sum^{n}_{r=1} r^2 = \frac{n(n + 1)(2n + 1)}{6}$$ $$\text{Second Series:}$$ $$\sum^{n}_{r=1} r = \frac{n(n + 1)}{2}$$3 $$\text{Third Series:}$$ $$\sum^{n}_{r=1} 1 = n

Part A - Step 3:

Now lets pop it into the equation $[r^2 + r - 1]$

r^2 = \frac{n(n + 1)(2n + 1)}{6}$$ $$r = \frac{n(n + 1)}{2}$$ $$-1 = (-1)n$$ $$\text{Adding them together: }$$ $$\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} + (-1) n$$ $$\text{Simpler, and making common denominators:}$$ $$\frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{6}\times3\times n(n + 1) - n$$ $$\text{(The second fraction is multiplied by 3 to}$$ $$\text{make the denominator 6, like the first one, and}$$ $$\text{is turned from } \frac{3}{6} \rightarrow \frac{1}{6} \times 3)

Part A - Step 4:

Now we can simplify fully by expanding brackets and algebraically manipulating:

\text{We have: }\frac{1}{6}n(n + 1)(2n + 1) + \frac{3}{6}n(n + 1) - n$$ $$\text{Factor out } \frac{1}{6}n:$$ $$\frac{1}{6}\,n\bigl[(n+1)(2n+1)+3(n+1)\bigr]\;-\;n.$$ $$\text{Factor } (n+1) \text{ out the square brackets:}$$ $$\frac{1}{6}\,n(n+1)[(2n+1)+3]\;-\;n.$$ $$\text{Simplify } (2n+1+3) \text{ to } (2n+4):$$ $$\frac{1}{6}\,n(n+1)(2n+4)\;-\;n.$$ $$\text{Write } -n \text{ as } -\frac{6}{6}n$$ $$\text{So we have: } \frac{1}{6}\,n(n+1)(2n+4)-\frac{6}{6}n

Part A - Step 5:

Further Simplification:

\text{We have: } \frac{1}{6}\,n(n+1)(2n+4)-\frac{6}{6}n$$ $$\text{Expand } (n+1)(2n+4) \text{ to } 2n^2 + 6n + 4$$ $$\text{So we have: } \frac{1}{6}\,n(2n^2 + 6n + 4)-\frac{6}{6}n$$ $$\text{Write as fractions: } \frac{n(2n^2 + 6n + 4)}{6} - \frac{6n}{6}$$ $$\text{Combine the fractions: } \frac{n(2n^2 + 6n + 4) - 6n}{6}$$ $$\text{Distribute: } n(2n^2 + 6n + 4) \text{ to } 2n^3 + 6n^2 + 4n$$ $$\text{So we have: } \frac{2n^3 + 6n^2 + 4n - 6n}{6}$$ $$\text{Simplify } 4n - 6n \text{ to } -2n$$ $$\text{So now we have: } \frac{2n^3 + 6n^2 - 2n}{6}$$ $$\text{Factor out: }2n \rightarrow \frac{2n(n^2 + 3n - 1)}{6}$$ $$\text{Simplify fractions: }$$ $$\frac{2n(n^2 + 3n - 1)}{6} \longrightarrow \frac{n(n^2 + 3n - 1)}{3}$$ $$\text{Final form is proved: }\boxed{ \sum^n_{r=1} [r^2+r-1] = \frac{n(n^2+3n-1)}{3}}

Part B - Step 1:

Now for part B, we can verify the result with n=3 and n=4. Lets start with n=3:

\text{We have: } \sum^n_{r=1} [r^2+r-1] = \frac{n(n^2+3n-1)}{3}$$ $$\text{Putting in } n=3:$$ $$\sum^3_{r=1} [r^2+r-1] = \frac{3(3^2+3\times3-1)}{3}$$ $$= \frac{3(9+9-1)}{3} = \frac{51}{3} = \boxed{17}

That was very long - no one else wouldve walked you though the algebra so clearly like that... I think it deserves a share 😉

⚠️ Common Mistakes

🚫 Mistake 1: Not splitting up their series'

What students often do wrong: Students may try to evaluate the entire series at once without breaking it down into manageable parts.

Why it's wrong: This approach can lead to mistakes in calculations and a lack of understanding of the underlying patterns in the series.

How to avoid it: Break the series down into smaller parts and evaluate each part separately. This will help you keep track of your calculations and understand the structure of the series better.

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Question 1

(A) Solve: $$ \sum_{r=1}^{99} r^2$$ (B) solve: $$\sum_{r=1}^{100} r^2$$ (C) Solve: $$\sum_{r=100}^{200} r^2$$ (D) Solve: $$\sum_{r=1}^{k} r^2 + \sum_{r=k+1}^{80} r^2,\quad 0 < k < 80$$

🔍 Show Solution

(A)
$$ \sum_{r=1}^{99} r^2 = \frac{99(99 + 1)(2\times99 + 1)}{6} = \frac{99\times100\times199}{6} = 328350$$ (B)
$$ \sum_{r=1}^{100} r^2 = \frac{100(100 + 1)(2\times100 + 1)}{6} = \frac{100\times101\times201}{6} = 338350$$ (C)
$$ \sum_{r=100}^{200} r^2 = \sum_{r=1}^{200} r^2 - \sum_{r=1}^{99} r^2$$ $$ = \frac{200(200 + 1)(2\times200 + 1)}{6} - 328350 = \frac{200\times201\times401}{6} - 328350 = 2378350$$ (D)
$$\text{This is really just: } \sum_{r=1}^{80} r^2\ \text{split into two parts.} $$ $$\text{This is because the first series has limit k, and the second has lower limit 1}$$ $$\text{more than k.}$$ $$\text{So, they add up to the same as } \sum_{r=1}^{80} r^2.$$ $$\quad \sum_{r=1}^{k} r^2 + \sum_{r=k+1}^{80} r^2 = \sum_{r=1}^{80} r^2 $$ $$\quad \text{Where} \quad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} $$ $$\quad \sum_{r=1}^{80} r^2 = \frac{80 \times 81 \times 161}{6} = \frac{1,\!044,\!960}{6} = 174,\!160$$

🎯 Practice Question 2

Let $$S(n) = \sum^n_{r=1} r^2, \quad H(n) = \sum^n_{r=1} r, \quad A(n) = \sum^n_{r=1} a \quad a \in \mathbb{Z} \quad r, n \in \mathbb{N}$$ (A) Given: $$T = \sum^n_{r=1}[3r^2 - 2r + 5], \text{ express } T \text{ in terms of } S(n), H(n) \text{ & } A(n)$$ (B) Hence, or otherwise, Find the exact value of $$\sum^{40}_{r=15} [3r^2 - 2r + 5]$$ (C) Find the value of: $$ n \text{ such that } S(n) = 1240$$ (D) Prove: $$\text{for any integer } m \geq 1: \sum^m_{r=1} [r^2 + (m - r)^2] = \frac{m(m + 1)(2m + 1)}{3}$$

🔍 Show Solution

```markdown (A)
$$T = \sum_{r=1}^{n} [3r^2 - 2r + 5]$$ $$3\sum_{r=1}^{n} r^2 - 2\sum_{r=1}^{n} r + \sum_{r=1}^{n} 5$$ $$3S(n) - 2H(n) + 5n$$ (B)
$$\sum_{r=15}^{40} [3r^2 - 2r + 5]$$ $$\sum_{r=1}^{40} [3r^2 - 2r + 5] - \sum_{r=1}^{14} [3r^2 - 2r + 5]$$ $$T_{40} - T_{14}$$ $$T_{40} = 3S(40) - 2H(40) + 5 \times 40$$ $$S(40) = \frac{40 \times 41 \times 81}{6} = 22140$$ $$H(40) = \frac{40 \times 41}{2} = 820$$ $$T_{40} = 3 \times 22140 - 2 \times 820 + 200 = 65220$$ $$T_{14} = 3S(14) - 2H(14) + 5 \times 14$$ $$S(14) = \frac{14 \times 15 \times 29}{6} = 1015$$ $$H(14) = \frac{14 \times 15}{2} = 105$$ $$T_{14} = 3 \times 1015 - 2 \times 105 + 70 = 2905$$ $$\sum_{r=15}^{40} [3r^2 - 2r + 5] = 65220 - 2905 = \boxed{62315}$$ (C)
$$S(n) = \frac{n(n+1)(2n+1)}{6}$$ $$\frac{n(n+1)(2n+1)}{6} = 1240$$ $$n(n+1)(2n+1) = 7440$$ $$n = 15 \quad \text{since } 15 \times 16 \times 31 = 7440$$ $$S(15) = \frac{15 \times 16 \times 31}{6} = 1240$$ $$\boxed{n = 15}$$ (D)
$$\sum_{r=1}^{m} [r^2 + (m - r)^2]$$ $$\sum_{r=1}^{m} r^2 + \sum_{r=1}^{m} (m - r)^2$$ $$\sum_{r=1}^{m} r^2 + \sum_{r=1}^{m} (m^2 - 2mr + r^2)$$ $$\sum_{r=1}^{m} r^2 + m^2 \cdot m - 2m \sum_{r=1}^{m} r + \sum_{r=1}^{m} r^2$$ $$2\sum_{r=1}^{m} r^2 - 2m \sum_{r=1}^{m} r + m^3$$ $$2S(m) - 2m H(m) + m^3$$ $$S(m) = \frac{m(m+1)(2m+1)}{6}, \quad H(m) = \frac{m(m+1)}{2}$$ $$2 \cdot \frac{m(m+1)(2m+1)}{6} - 2m \cdot \frac{m(m+1)}{2} + m^3$$ $$\frac{m(m+1)(2m+1)}{3} - m^2(m+1) + m^3$$ $$\frac{m(m+1)(2m+1)}{3} - m^2$$ $$\boxed{\sum_{r=1}^{m} [r^2 + (m - r)^2] = \frac{m(m+1)(2m+1)}{3} - m^2}$$

📋 Summary

🎯 Key Takeaways

Understanding the properties of square numbers and their sums.
Application of summation formulas to solve complex series problems.
Proving identities involving sums of squares and linear terms.

📚 What's Next?

Further your understanding and knowledge by doing practice questions, and I suggest doing my next lesson on Cubic Series, which builds on this lesson.
You can find that lesson here.

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