P1 §8.1 Pascal's triangle

Pure mathematics Year 1

Table of contents

1. Binomial, trinomial and polynomial
2. Binomial expansion
3. Pascal's triangle
4. Examples
5. Edexcel P1 Ch8 Exercise 8A

1. Binomial, trinomial and polynomial

The binomial expansion tells us how to expand $(a+b)^n$.

 

The word 'bi-nomial' means 'two terms' - here, they are $a$ and $b$. Similarly, monomial means 'one term', trinomial means 'three terms' and polynomial means 'many terms' and so on.

 

• The 'trinomial expansion' would concern $(a+b+c)^n$ and the 'polynomial expansion' would $(a_1+a_2+\cdots+a_k)^n$, but they do not require a separate treatment as they are in principle built upon the binomial case. For example, the trinomial expansion is just two successive applications of the binomial expansion : $$\begin{align} (a+b+c)^n=[a+(b+c)]^n\end{align}$$

2. Binomial expansion

We consider the following examples which can be expanded by hand. (If you have not done this calculation before, it is highly recommended to do so at least once in your lifetime.) $$\begin{align} (a+b)^0&=1 \\ (a+b)^1&=a+b \\ (a+b)^2&=a^2+2ab+b^2 \\ (a+b)^3&=a^3+3a^2b+3ab^2+b^3 \\ (a+b)^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4 \\ (a+b)^5&=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 \\ &\;\;\vdots \end{align}$$

 

• Every term in the expansion of $(a+b)^n$ has total index/power $n$. For example, in the expansion of $(a+b)^4$, we have $a^4,a^3b,a^2b^2,ab^3,b^4$ terms and the total index of each term is 4.

3. Pascal's triangle

The coefficients in the expansions form a pattern that is known as Pascal's triangle, as shown below. $$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ \\ (n=1)&&&&&&& 1 && 1 \\ \\ (n=2)&&&&&& 1 && 2 && 1 \\ \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&&&&&&\vdots \end{array}$$

 

• Pascal's triangle is formed by adding adjacent pairs of numbers to find the numbers on the next row.
• The $(n+1)$th row of Pascal's triangle gives the coefficients in the expansions of $(a+b)^n$.

4. Examples

Example 1. Use Pascal's triangle to find the expansions of: $$\begin{align} {\rm (a)}&&&(x+2y)^3 \\ {\rm (b)}&&&(2x-5)^4 \end{align}$$

 

(Edexcel 2017 Specifications, P1 Ch8 Example 1.)

 

 

Solution.

Recall Pascal's triangle:

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$

 

(a) The binomial coefficients are (Index $=3$ so we look at the 4th row of Pascal's triangle to find the coefficnets.)

$$\begin{align} 1 \quad 3 \quad 3 \quad 1 \end{align}$$ We have ($a=x$ and $b=2y$) $$\begin{align} (a+b)^3&=a^3+3a^2b+3ab^2+b^3 \\ \\ \Rightarrow\quad (x+2y)^3 &=x^3+3x^2(2y)+3x(2y)^2+(2y)^3 \\ &=x^3+6x^2y+12xy^2+8y^3 \end{align}$$

 

(b) The binomial coefficients are: (The index $=4$ so we look at the 5th row of Pascal's triangle to find the coefficients.)

$$\begin{align} 1 \quad 4 \quad 6 \quad 4 \quad 1 \end{align}$$ We have ($a=2x$ and $b=-5$) $$\begin{align} (a+b)^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4 \\ \\ \Rightarrow\quad (2x-5)^4 &=(2x)^4+4(2x)^3(-5)+6(2x)^2(-5)^2+4(2x)(-5)^3+(-5)^4 \\ &=16x^4-160x^3+600x^2-1000x+625 \end{align}$$

 

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Example 2. The coefficient of $x^2$ in the expansion of $(2-cx)^3$ is 294. Find the possible value(s) of the constant $c$.

 

(Edexcel 2017 Specifications, P1 Ch8 Example 2.)

 

Solution.

Recall Pascal's triangle:

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$

 

The binomial coefficients are: (The index $=3$ so we look at the 4th row of Pascal's triangle to find the coefficients.) $$\begin{align} 1 \quad 3 \quad 3 \quad 1 \end{align}$$

The $x^2$ term is given by (from the expansion of $(a+b)^3$ where $a=2$ and $b=-cx$) $$\begin{align} 3\times2(-cx)^2=6c^2x^2 \end{align}$$ and we find $$\begin{align} 6c^2&=294 \\ \Rightarrow\quad c^2&=49 \\ \Rightarrow\quad c&=\pm7 \quad\checkmark \end{align}$$

 

Aside. We can of course fully expand $(2-cx)^3$: $$\begin{align} (a+b)^3&=a^3+3a^2b+3ab^2+b^3 \\ \\ \Rightarrow\quad (2-cx)^3 &=2^3+3\times2^2(-cx)+3\times2(-cx)^2+(-cx)^3 \\ &=8-12cx+6c^2x^2-c^3x^3 \end{align}$$

5. Edexcel P1 Ch8 Exercise 8A

Question 1. State the row of Pascal's triangle that would give the coefficients of each expansion: $$\begin{align} {\rm (a)}&&& (x+y)^3 \\ {\rm (b)}&&& (3x-7)^{15} \\ {\rm (c)}&&& \left(2x+\frac12\right)^n \\ {\rm (d)}&&& (y-2x)^{n+4} \end{align}$$

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q1.)

 

Solution.

Recall Pascal's triangle:

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$

 

(a) The index is 3 so the 4th row of Pascal's triangle provides the binomial coefficients: $$\begin{align} 1 \quad 3 \quad 3 \quad 1 \end{align}$$ Although not asked, for completeness, we find: $$\begin{align} (x+y)^3&=x^3+3x^2+y+3xy^2+y^3 \end{align}$$

 

(b) The index is 15 so the 16th row of Pascal's triangle provides the binomial coefficients: $$\begin{align} 1 \quad 15 \quad 105 \quad 455 \quad 1365 \quad 3003 \quad 5005 \quad 6435 \quad 6435 \quad 5005 \quad 3003 \quad 1365 \quad 455 \quad 105 \quad 15 \quad 1 \end{align}$$ Although not asked, for completeness, we find: $$\begin{align} (3x-7)^{15} &={\color{blue}{(3x)^{15}}} +15{\color{blue}{(3x)^{14}}}{\color{purple}{(-7)}} +105{\color{blue}{(3x)^{13}}}{\color{purple}{(-7)^{2}}} +455{\color{blue}{(3x)^{12}}}{\color{purple}{(-7)^{3}}} +1365{\color{blue}{(3x)^{11}}}{\color{purple}{(-7)^{4}}} \\ &\qquad +3003{\color{blue}{(3x)^{10}}}{\color{purple}{(-7)^{5}}} +5005{\color{blue}{(3x)^{9}}}{\color{purple}{(-7)^{6}}} +6435{\color{blue}{(3x)^{8}}}{\color{purple}{(-7)^{7}}} +6435{\color{blue}{(3x)^{7}}}{\color{purple}{(-7)^{8}}} +5005{\color{blue}{(3x)^{6}}}{\color{purple}{(-7)^{9}}} \\ &\qquad +3003{\color{blue}{(3x)^{5}}}{\color{purple}{(-7)^{10}}} +1365{\color{blue}{(3x)^{4}}}{\color{purple}{(-7)^{11}}} +455{\color{blue}{(3x)^{3}}}{\color{purple}{(-7)^{12}}} +105{\color{blue}{(3x)^{2}}}{\color{purple}{(-7)^{13}}} +15{\color{blue}{(3x)}}{\color{purple}{(-7)^{14}}} +{\color{purple}{(-7)^{15}}} \\ &= 14348907 x^{15} - 502211745 x^{14} + 8202791835 x^{13} - 82939339665 x^{12} + 580575377655 x^{11} \\ &\qquad - 2980286938629 x^{10} + 11590004761335 x^9 - 34770014284005 x^8 + 81130033329345 x^7 - 147235986412515 x^6 \\ &\qquad + 206130380977521 x^5 - 218623131339795 x^4 + 170040213264285 x^3 - 91560114834615 x^2 + 30520038278205 x \\ &\qquad - 4747561509943 \end{align}$$

 

(c) The index is $n$ so the ($n+1$)-th row of Pascal's triangle provides the binomial coefficients.

 

(d) The index is $n+4$ so the ($n+5$)-th row of Pascal's triangle provides the binomial coefficients.

 

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Question 2. Write down the expansion of: $$\begin{align} {\rm (a)}&&& (x+y)^4 \\ {\rm (b)}&&& (p+q)^5 \\ {\rm (c)}&&& (a-b)^3 \\ {\rm (d)}&&& (x+4)^3 \\ {\rm (e)}&&& (2x-3)^4 \\ {\rm (f)}&&& (a+2)^5 \\ {\rm (g)}&&& (3x-4)^4 \\ {\rm (h)}&&& (2x-3y)^4 \end{align}$$

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q2.)

 

Solution.

Recall Pascal's triangle:

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$

 

$$\begin{align} {\textbf{(a)}}&& (x+y)^4&={\color{blue} 1}x^4 + {\color{blue} 4}x^3y + {\color{blue} 6}x^2y^2 + {\color{blue} 4}xy^3 + {\color{blue} 1}y^4 \\ \\ {\textbf{(b)}}&& (p+q)^5&={\color{blue} 1}p^5 + {\color{blue} 5}p^4q + {\color{blue}{10}}p^3q^2 + {\color{blue}{10}}p^2q^3 + {\color{blue} 5}pq^4 + {\color{blue} 1}q^5 \\ \\ {\textbf{(c)}}&& (a-b)^3 &={\color{blue} 1}\cdot a^3 + {\color{blue} 3}a^2(-b) + {\color{blue} 3}a(-b)^2 + {\color{blue} 1}(-b)^3 \\ &&&=a^3-3a^2b+3ab^2-b^3 \\ \\ {\textbf{(d)}}&& (x+4)^3 &={\color{blue} 1}\cdot x^3 + {\color{blue} 3}x^2\cdot4 + {\color{blue} 3}x\cdot4^2 + {\color{blue} 1}\cdot4^3 \\ &&&=x^3+12x^2+48x+64 \\ \\ {\textbf{(e)}}&& (2x-3)^4 &={\color{blue} 1}\cdot(2x)^4 + {\color{blue} 4}(2x)^3(-3) + {\color{blue} 6}(2x)^2(-3)^2 + {\color{blue} 4}(2x)(-3)^3 + {\color{blue} 1}\cdot(-3)^4 \\ &&&=16 x^4 - 96 x^3 + 216 x^2 - 216 x + 81 \\ \\ {\textbf{(f)}}&& (a+2)^5 &={\color{blue} 1}\cdot a^5 + {\color{blue} 5}a^4\cdot 2 + {\color{blue} {10}}a^3\cdot 2^2 + {\color{blue} {10}}a^2\cdot 2^3 + {\color{blue} 5}a\cdot 2^4 + {\color{blue} 1}\cdot 2^5 \\ &&&=a^5 + 10 a^4 + 40 a^3 + 80 a^2 + 80 a + 32 \\ \\ {\textbf{(g)}}&& (3x-4)^4 &={\color{blue} 1}\cdot(3x)^4 + {\color{blue} 4}(3x)^3(-4) + {\color{blue} 6}(3x)^2(-4)^2 + {\color{blue} 4}(3x)(-4)^3 + {\color{blue} 1}\cdot(-4)^4 \\ &&&=81 x^4 - 432 x^3 + 864 x^2 - 768 x + 256 \\ \\ {\textbf{(h)}}&& (2x-3y)^4 &={\color{blue} 1}(2x)^4 + {\color{blue} 4}(2x)^3(-3y) + {\color{blue} 6}(2x)^2(-3y)^2 + {\color{blue} 4}(2x)(-3y)^3 + {\color{blue} 1}(-3y)^4 \\ &&&=16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4 \end{align}$$

 

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Question 3. Find the coefficient of $x^3$ in the expansion of: $$\begin{align} {\rm (a)}&&& (4+x)^4 \\ {\rm (b)}&&& (1-x)^5 \\ {\rm (c)}&&& (3+2x)^3 \\ {\rm (d)}&&& (4+2x)^5 \\ {\rm (e)}&&& (2+x)^6 \\ {\rm (f)}&&& \left(4-\frac12x\right)^4 \\ {\rm (g)}&&& (x+2)^5 \\ {\rm (h)}&&& (3-2x)^4 \end{align}$$

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q3.)

 

Solution.

$$\begin{align} {\textbf{(a)}}&& &(4+x)^4&&\Rightarrow& {\color{blue} 4}\cdot 4x^3&=16x^3 \\ \\ {\textbf{(b)}}&& &(1-x)^5&&\Rightarrow& {\color{blue}{10}}(-x)^3&=-10x^3 \\ \\ {\textbf{(c)}}&& &(3+2x)^3&&\Rightarrow& {\color{blue} 1}(2x)^3&=8x^3 \\ \\ {\textbf{(d)}}&& &(4+2x)^5&&\Rightarrow& {\color{blue} {10}}\cdot4^2 (2x)^3&=1280x^3 \\ \\ {\textbf{(e)}}&& &(2+x)^6&&\Rightarrow& {\color{blue} 20}\cdot 2^3x^3&=160x^3 \\ \\ {\textbf{(f)}}&& &\left(4-\frac12x\right)^4&&\Rightarrow& {\color{blue} 4}\cdot4\left(-\frac12x\right)^3&=-2x^3 \\ \\ {\textbf{(g)}}&& &(x+2)^5&&\Rightarrow& {\color{blue} 10}x^3\cdot 2^2&=40x^3 \\ \\ {\textbf{(h)}}&& &(3-2x)^4&&\Rightarrow& {\color{blue} 4}\cdot 3(-2x)^3&=-96x^3 \end{align}$$

 

Aside. The full expansions read: $$\begin{align} {\textbf{(a)}}&& (4+x)^4&={\color{blue} 1}\cdot 4^4 + {\color{blue} 4}\cdot 4^3x + {\color{blue} 6}\cdot 4^2x^2 + {\color{blue} 4}\cdot 4x^3 + {\color{blue} 1}x^4 \\ &&&=256 + 256 x + 96x^2 + 16 x^3 + x^4 \\ \\ {\textbf{(b)}}&& (1-x)^5&={\color{blue} 1} + {\color{blue} 5}(-x) + {\color{blue}{10}}(-x)^2 + {\color{blue}{10}}(-x)^3 + {\color{blue} 5}(-x)^4 + {\color{blue} 1}(-x)^5 \\ &&&=1 - 5 x + 10 x^2 - 10 x^3 + 5 x^4 - x^5 \\ \\ {\textbf{(c)}}&& (3+2x)^3 &={\color{blue} 1}\cdot 3^3 + {\color{blue} 3}\cdot 3^2(2x) + {\color{blue} 3}\cdot 3(2x)^2 + {\color{blue} 1}(2x)^3 \\ &&&=27 + 54 x + 36 x^2 + 8 x^3 \\ \\ {\textbf{(d)}}&& (4+2x)^5 &={\color{blue} 1}\cdot 4^5 + {\color{blue} 5}\cdot 4^4(2x) + {\color{blue} {10}}\cdot 4^3(2x)^2 + {\color{blue} {10}}\cdot 4^2(2x)^3 + {\color{blue} 5}\cdot 4(2x)^4 + {\color{blue} 1}\cdot(2x)^5 \\ &&&=1024 + 2560 x + 2560 x^2 + 1280 x^3 + 320 x^4 + 32 x^5 \\ \\ {\textbf{(e)}}&& (2+x)^6 &={\color{blue} 1}\cdot 2^6 + {\color{blue} 6}\cdot 2^5x + {\color{blue} 15}\cdot 2^4x^2 + {\color{blue} 20}\cdot 2^3x^3 + {\color{blue} 15}\cdot 2^2x^4 + {\color{blue} 6}\cdot 2x^5 + {\color{blue} 1}x^6 \\ &&&=64 + 192 x + 240 x^2 + 160 x^3 + 60 x^4 + 12 x^5 + x^6 \\ \\ {\textbf{(f)}}&& \left(4-\frac12x\right)^4 &={\color{blue} 1}\cdot 4^4 + {\color{blue} 4}\cdot 4^3\left(-\frac12x\right) + {\color{blue} 6}\cdot 4^2\left(-\frac12x\right)^2 + {\color{blue} 4}\cdot 4\left(-\frac12x\right)^3 + {\color{blue} 1}\cdot \left(-\frac12x\right)^4 \\ &&&=256 - 128 x + 24 x^2 - 2 x^3 + x^4/16 \\ \\ {\textbf{(g)}}&& (x+2)^5 &={\color{blue} 1}\cdot x^5 + {\color{blue} 5}x^4\cdot 2 + {\color{blue} {10}}x^3\cdot 2^2 + {\color{blue} {10}}x^2\cdot 2^3 + {\color{blue} 5}x\cdot 2^4 + {\color{blue} 1}\cdot 2^5 \\ &&&=x^5 + 10 x^4 + 40 x^3 + 80 x^2 + 80 x + 32 \\ \\ {\textbf{(h)}}&& (3-2x)^4 &={\color{blue} 1}\cdot 3^4 + {\color{blue} 4}\cdot 3^3(-2x) + {\color{blue} 6}\cdot 3^2(-2x)^2 + {\color{blue} 4}\cdot 3(-2x)^3 + {\color{blue} 1}(-2x)^4 \\ &&&=81 - 216 x + 216 x^2 - 96 x^3 + 16 x^4 \end{align}$$

 

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Question 4. (P) Fully expand the expression $$\begin{align} (1+3x)(1+2x)^3 \end{align}$$

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q4.)

 

Solution.

We expand $(1+2x)^3$, then multiply it by $(1+3x)$. Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ we find: $$\begin{align} (1+2x)^3 &={\color{blue} 1} + {\color{blue} 3}(2x) + {\color{blue} 3}(2x)^2 + {\color{blue} 1}(2x)^3 \\ &=1 + 6 x + 12 x^2 + 8 x^3 \end{align}$$ and then: $$\begin{align} (1+3x)(1+2x)^3 &=(1+3x)\left( 1 + 6 x + 12 x^2 + 8 x^3 \right) \\ &=1 + 6 x + 12 x^2 + 8 x^3 \\ &\qquad+ 3x + 18x^2 + 36x^3 + 24x^4 \\ &=1 + 9x + 30 x^2 + 44 x^3 + 24 x^4 \qquad\checkmark \end{align}$$

 

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Question 5. (P) Expand $(2+y)^3$. Hence or otherwise, write down the expansion of $\left(2+x-x^2\right)^3$ in ascending powers of $x$.

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q5.)

 

Solution.

Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ we find: $$\begin{align} (2+y)^3 &={\color{blue} 1}\cdot 2^3 + {\color{blue} 3}\cdot 2^2y + {\color{blue} 3}\cdot 2 y^2 + {\color{blue} 1}\cdot y^3 \\ &=8 + 12 y + 6 y^2 + y^3 \\ \\ \Rightarrow\qquad \left(2+x-x^2\right)^3 &=8 + 12 \left(x-x^2\right) + 6 \left(x-x^2\right)^2 + \left(x-x^2\right)^3 \\ &=8 + 12 \left(x-x^2\right) + 6 \left(x^2 -2x^3 + x^4\right) + \left[ {\color{blue} 1}\cdot x^3 + {\color{blue} 3}x^2\left(-x^2\right) + {\color{blue} 3}x\left(-x^2\right)^2 + {\color{blue} 1}\left(-x^2\right)^3 \right] \\ &=8 + 12 \left(x-x^2\right) + 6 \left(x^2 -2x^3 + x^4\right) + \left( x^3 - 3 x^4 + 3 x^5 - x^6 \right) \\ &=8 + 12x + (-12 + 6)x^2 + (-12 +1) x^3 + (6 - 3) x^4 + 3x^5 - x^6 \\ &=8 + 12x - 6x^2 - 11 x^3 + 3 x^4 + 3 x^5 - x^6 \qquad\checkmark \end{align}$$

 

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Question 6. (P) The coefficient of $x^2$ in the expansion of $(2+ax)^3$ is 54. Find the possible values of the constant $a$.

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q6.)

 

Solution.

Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ we find (of course, we only need the $x^2$ term): $$\begin{align} (2+ax)^3 &={\color{blue} 1}\cdot 2^3 + {\color{blue} 3}\cdot 2^2(ax) + {\color{blue} 3}\cdot 2 (ax)^2 + {\color{blue} 1}\cdot (ax)^3 \\ &=8 + 12ax + 6a^2x^2 + a^3x^3 \end{align}$$ It is given that the coefficient of $x^2$ is 54, that is: $$\begin{align} && 6a^2&=54 \\ &\Rightarrow& a^2&=9 \\ &\Rightarrow& a&=\pm 3 \qquad\checkmark \end{align}$$

 

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Question 7. (P) The coefficient of $x^2$ in the expansion of $$\begin{align} (2-x)(3+bx)^3 \end{align}$$ is 45. Find possible values of the constant $b$.

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q7.)

 

Solution.

Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ we find (of course, we only need the $x^2$ term): $$\begin{align} (3+bx)^3 &={\color{blue} 1}\cdot 3^3 + {\color{blue} 3}\cdot 3^2(bx) + {\color{blue} 3}\cdot 3 (bx)^2 + {\color{blue} 1}\cdot (bx)^3 \\ &=27 + 27bx + 9b^2x^2 + b^3x^3 \\ \\ \Rightarrow\qquad (2-x)(3+bx)^3 &=(2-x)\left(27 + 27bx + 9b^2x^2 + b^3x^3\right) \\ &=\left(54 + 54bx + 18b^2x^2 + 2b^3x^3\right) - \left(27x + 27bx^2 + 9b^2x^3 + b^3x^4\right) \\ &=54 + (54b - 27)x + \left( 18b^2 - 27b \right)x^2 + \left( 2b^3 - 9b^2 \right) x^3 - b^3x^4 \\ &=54 + 27(2b-1)x + 9b \left( 2b - 3 \right)x^2 + b^2\left( 2b - 9 \right) x^3 - b^3x^4 \\ \end{align}$$ It is given that the coefficient of $x^2$ is 45 (so, strictly speaking, we only need the $x^2$ term above), that is: $$\begin{align} && 9b \left( 2b - 3 \right)&=45 \\ &\Rightarrow& 2b^2 - 3b &= 5 \\ &\Rightarrow& 2b^2 - 3b - 5 &= 0 \\ &\Rightarrow& (2b - 5)(b + 1)&=0 \\ &\Rightarrow& b=\frac52\quad{\textrm{or}}\quad b&=-1 \qquad\checkmark \end{align}$$

 

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Question 8. (P) Work out the coefficient of $x^2$ in the expansion of $$\begin{align} (p-2x)^3 \end{align}$$ Given your answer in terms of $p$.

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q8.)

 

Solution.

Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ we find (of course, we only need the $x^2$ term): $$\begin{align} (p-2x)^3 &={\color{blue} 1}\cdot p^3 + {\color{blue} 3}p^2(-2x) + {\color{blue} 3}p (-2x)^2 + {\color{blue} 1}\cdot (-2x)^3 \\ &=p^3 - 6p^2x + 12px^2 - 8x^3 \end{align}$$ So the coefficient of $x^2$ is $12p.\qquad\checkmark$

 

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Question 9. (P) After 5 years, the value of an investment of $500$ at an interest rate of $X\%$ per annum is given by: $$\begin{align} 500\left(1+\frac{X}{100}\right)^5 \end{align}$$ Find an approximation for this expression in the form $$\begin{align} A+BX+CX^2, \end{align}$$ where $A,B$ and $C$ are constants to be found. You can ignore higher powers of $X$.

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Q9.)

 

Solution.

Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ we find: $$\begin{align} \left(1+\frac{X}{100}\right)^5 &={\color{blue} 1} + {\color{blue} 5}\left(\frac{X}{100}\right) + {\color{blue} {10}}\left(\frac{X}{100}\right)^2 + {\color{blue} {10}}\left(\frac{X}{100}\right)^3 + {\color{blue} 5}\left(\frac{X}{100}\right)^4 + {\color{blue} 1}\left(\frac{X}{100}\right)^5 \\ &=1+\frac{5X}{100} + \frac{10X^2}{10000} + \cdots \\ &=1+\frac{X}{20} + \frac{X^2}{1000} + \cdots \\ \\ \Rightarrow\qquad 500\left(1+\frac{X}{100}\right)^5 &=500\left( 1+\frac{X}{20} + \frac{X^2}{1000} + \cdots \right) \\ &=500 + \frac{500X}{20} + \frac{500X^2}{1000} + \cdots \\ &=500 + 25X + \frac{X^2}{2} + \cdots \qquad\checkmark \end{align}$$

 

Aside. A derivation of compound interest: With an initial investment $a$ at an interest rate of $X\%$, we find: $$\begin{align} {\textrm{After 1 year:}}&& A_1&=a + a\times\frac{X}{100} = a\left( 1+\frac{X}{100} \right) \\ {\textrm{After 2 years:}}&& A_2&=A_1 + A_1\times\frac{X}{100} = A_1\left( 1+\frac{X}{100} \right)=a\left( 1+\frac{X}{100} \right)^2 \\ {\textrm{After 3 years:}}&& A_3&=A_2 + A_2\times\frac{X}{100} = A_2\left( 1+\frac{X}{100} \right)=a\left( 1+\frac{X}{100} \right)^3 \\ &&&\vdots \\ {\textrm{After $n$ years:}}&& A_n&=A_{n-1} + A_{n-1}\times\frac{X}{100} = A_{n-1}\left( 1+\frac{X}{100} \right)=a\left( 1+\frac{X}{100} \right)^n \qquad\square \end{align}$$

 

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Challenge. Find the constant term in the expansion of $$\begin{align} \left(x^2-\frac1{2x}\right)^3. \end{align}$$

 

(Edexcel 2017 Specifications, P1 Ch8 Exercise 8A Challenge.)

 

Solution.

Using Pascal's triangle,

$$\begin{array}{ccccccccccccccc} (n=0)&&&&&&&& 1 \\ (n=1)&&&&&&& 1 && 1 \\ (n=2)&&&&&& 1 && 2 && 1 \\ (n=3)&&&&& 1 && 3 && 3 && 1 \\ (n=4)&&&& 1 && 4 && 6 && 4 && 1 \\ (n=5)&&&1 && 5 && 10 && 10 && 5 && 1 \\ &&&& && && \vdots && && \end{array}$$ the constant term reads: $$\begin{align} \left(x^2-\frac1{2x}\right)^3 &={\color{blue} 1}\left( x^2 \right)^3 + {\color{blue} 3}\left( x^2 \right)^2\left(-\frac1{2x}\right) + \underbrace{{\color{blue} 3}\left( x^2 \right)\left(-\frac1{2x}\right)^2}_{=3x^2 \times \frac1{4x^2}=\frac34\quad\checkmark} + {\color{blue} 1}\left(-\frac1{2x}\right)^3 \end{align}$$

 

Aside. The full expansion reads: $$\begin{align} \left(x^2-\frac1{2x}\right)^3 &={\color{blue} 1}\left( x^2 \right)^3 + {\color{blue} 3}\left( x^2 \right)^2\left(-\frac1{2x}\right) + {\color{blue} 3}\left( x^2 \right)\left(-\frac1{2x}\right)^2 + {\color{blue} 1}\left(-\frac1{2x}\right)^3 \\ &=x^6 - \frac{3 x^3}{2} +\frac34 - \frac1{8x^3} \end{align}$$