Formulas
Binomial expansion (provided):
\begin{align}
(a+b)^n & =a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+...+\binom{n}{r}a^{n-r}b^r+ \ldots + b^n \\
\\
\text{where } n \text{ is a positive integer and }&\binom{n}{r}={n!\over r!(n-r)!}={n(n-1) \ldots (n-r+1)\over r!}
\end{align}
Note: Use the nCr function in calculator to find the value of ${n \choose r}$.
Binomial coefficients with n:
${n \choose 1}$ = $ \phantom{.} n $
${n \choose 2}$ = $ {n(n - 1) \over 2!} $
${n \choose 3}$ = $ {n(n - 1)(n - 2) \over 3!} $
Proof (Method 1)
\begin{align}
\require{cancel}
\binom{n}{r} & = {n! \over r! (n - r)! }
\phantom{000000} [\text{In formula sheet}] \\
\\
\binom{n}{1} & = {n! \over 1! (n - 1)!} \\
& = {n \times \cancel{(n - 1) \times ... \times 2 \times 1} \over (1) [ \cancel{(n - 1) \times ... \times 2 \times 1!}] } \\
& = {n \over 1} \\
& = n \\
\\
\binom{n}{2} & = {n! \over 2! (n - 2)!} \\
& = {n \times (n - 1) \times \cancel{ (n - 2) \times ... \times 2 \times 1} \over 2! [ \cancel{ (n - 2) \times ... \times 2 \times 1 } ] } \\
& = {n \times (n - 1) \over 2!} \\
& = {n(n - 1) \over 2!} \\
\\
\binom{n}{3} & = {n! \over 3! (n - 3)!} \\
& = {n \times (n - 1) \times (n - 2) \times \cancel{ (n - 3) \times ... \times 2 \times 1 } \over 3![ \cancel{ (n - 3) \times ... \times 2 \times 1} ] } \\
& = {n (n - 1) (n - 2) \over 3!}
\end{align}
Proof (Method 2)
\begin{align}
\require{cancel}
\binom{n}{r} & = {\overbrace{n(n - 1)...(n - r + 1)}^{r \text{ factors}} \over r!}
\phantom{000000} [\text{In formula sheet}] \\
\\
\binom{n}{1} & = {n \over 1!} \\
& = {n \over 1} \\
& = n \\
\\
\binom{n}{2} & = {n(n - 1) \over 2!} \\
\\
\binom{n}{3} & = {n(n - 1)(n - 2) \over 3!} \\
\end{align}
Questions
Expansion questions
1(i) Find, in ascending powers of $x$, the first three terms in the expansion of $(1 + 2x)^5$.
Answer: $ 1 + 10x + 40x^2 + ... $
Solutions
\begin{align}
(1 + 2x)^5 & = 1^5 + {5 \choose 1} (1)^4 (2x) + {5 \choose 2} (1)^3 (2x)^2 + ... \\
& = 1 + (5)(1)(2x) + (10)(1)(4x^2) + ... \\
& = 1 + 10x + 40x^2 + ...
\end{align}
1(ii) Find, in ascending powers of $x$, the first three terms in the expansion of $(3 - 4x)^5$.
Answer: $ 243 - 1620x + 4320x^2 + ... $
Solutions
\begin{align}
(3 - 4x)^5 & = 3^5 + {5 \choose 1} (3)^4 (-4x) + {5 \choose 2} (3)^3 (-4x)^2 + ... \\
& = 243 + (5)(81)(-4x) + (10)(27)(16x^2) + ... \\
& = 243 - 1620x + 4320x^2 + ...
\end{align}
1(iii) Using your answers from (i) and (ii), find the coefficient of $x^2$ in the expansion of $(3 + 2x - 8x^2)^5$.
Answer: $ -2160 $
Solutions
\begin{align}
3 + 2x - 8x^2 & = (1 + 2x)(3 - 4x) \\
\\
\therefore (3 + 2x - 8x^2)^5 & = [(1 + 2x)(3 - 4x)]^5 \\
& = (1 + 2x)^5 (3 - 4x)^5 \\
& = (1 + 10x + 40x^2 + ...)(243 - 1620x + 4320x^2 + ...) \\
& = ... + (1)(4320x^2) + ... + (10x)(-1620x) + ... + (40x^2)(243) + ...
\phantom{000000} [\text{Only need terms in } x^2] \\
& = 4320x^2 - 16 \phantom{.} 200x + 9720x^2 + ... \\
& = -2160x^2 + ... \\
\\
\text{Coefficient of } & x^2 = - 2160
\end{align}
2. When $(1 - x)(1 + ax)^6$ is expanded as far as the term in $x^2$, the result is $1 + bx^2$. Find the value of $a$ and of $b$.
Hint: The coefficient of $x$ in the expansion of $(1 - x)(1 + ax)^6$ is $ 0 $ since it does not exist
(from A Maths 360 2nd edition Ex 6.1)
Answer: $ a = {1 \over 6}, b = -{7 \over 12} $
Solutions
\begin{align}
(1 + ax)^6 & = (1)^6 + \binom{6}{1} (1)^5 (ax)^1 + \binom{6}{2} (1)^4 (ax)^2 + ... \\
& = 1 + (6)(1)(ax) + (15)(1)(a^2x^2) + ... \\
& = 1 + 6ax + 15a^2 x^2 + ... \\
\\
(1 - x)(1 + ax)^6 & = (1 - x)(1 + 6ax + 15a^2x^2 + ...) \\
& = (1)(1) + (1)(6ax) + (1)(15a^2x^2) + (-x)(1) + (-x)(6ax) + ... \phantom{00000} [\text{Ignore terms higher than } x^2] \\
& = 1 + 6ax + 15a^2x^2 - x - 6ax^2 + ... \\
& = 1 + 6ax - x + 15a^2x^2 - 6ax^2 + ... \\
& = 1 + (6a - 1)x + (15a^2 - 6a)x^2 + ... \\
\\
\therefore 1 + bx^2 & = 1 + (6a - 1)x + (15a^2 - 6a)x^2 + ... \\
1 + 0x + bx^2 & = 1 + (6a - 1)x + (15a^2 - 6a)x^2 + ... \\
\\
\text{Comparing } & \text{coefficient of } x, \\
0 & = 6a - 1 \\
1 & = 6a \\
{1 \over 6} & = a \\
\\
\text{Comparing } & \text{coefficient of } x^2, \\
b & = 15a^2 - 6a \\
b & = 15 \left(1 \over 6\right)^2 - 6\left(1 \over 6\right) \phantom{00000000} \left[ \text{Substitute } a = {1 \over 6} \right] \\
b & = -{7 \over 12}
\end{align}
Substitution question
3(i) Find, in ascending powers of $k$, the first four terms in the expansion of $(2 + k)^8 $.
Answer: $ 256 + 1024k + 1792k^2 + 1792k^3 + ... $
Solutions
\begin{align}
(2 + k)^8 & = 2^8 + {8 \choose 1} (2)^7 (k) + {8 \choose 2} (2)^6 (k)^2 + {8 \choose 3} (2)^5 (k)^3 + ... \\
& = 256 + (8)(128)(k) + (28)(64)(k^2) + (56)(32)(k^3) + ... \\
& = 256 + 1024k + 1792k^2 + 1792k^3 + ...
\end{align}
3(ii) Hence, by using a suitable substitution, find the first four terms in the expansion of $(2 + x^2 - x)^8$.
Answer: $ 256 - 1024x + 2816x^2 - 5366x^3 + ... $
Solutions
\begin{align}
\text{From (i), } (2 + k)^8 & = 256 + 1024k + 1792k^2 + 1792k^3 + ... \\
\\
\text{Replace } & k \text{ by } x^2 - x, \\
(2 + x^2 - x)^8 & = 256 + 1024 (x^2 - x) + 1792 (x^2 - x)^2 + 1792 (x^2 - x)^3 + ... \\
& = 256 + 1024x^2 - 1024x + 1792[ \underbrace{ (x^2)^2 - 2(x^2)(x) + x^2 }_{(a - b)^2 = a^2 - 2ab + b^2}]
+ 1792(x^2 - x)^2 (x^2 - x) + ... \\
& = 256 + 1024x^2 - 1024x + 1792 (x^4 - 2x^3 + x^2) + 1792(x^4 - 2x^3 + x^2)(x^2 - x) + ... \\
& = 256 + 1024x^2 - 1024x + ... - 3584x^3 + 1792x^2 + 1792(... - x^3) + ...
\phantom{000000} [\text{Only first 4 terms - till } x^3] \\
& = 256 - 1024x + 2816x^2 - 3584x^3 - 1792x^3 + ... \\
& = 256 - 1024x + 2816x^2 - 5376x^3 + ...
\end{align}
Approximation question
4(i) Find, in ascending powers of $x$, the first four terms in the expansion of $\left( 1 - {1 \over 2}x\right)^7$.
Answer: $ 1 - {7 \over 2}x + {21 \over 4}x^2 - {35 \over 8}x^3 + ... $
Solutions
\begin{align}
\left(1 - {1 \over 2}x\right)^7 & = 1^7 + {7 \choose 1} (1)^6 \left(-{1 \over 2}x\right) + {7 \choose 2} (1)^5 \left(-{1 \over 2}x\right)^2 + {7 \choose 3} (1)^4 \left(-{1 \over 2}x\right)^3 + ... \\
& = 1 + (7)(1)\left(-{1 \over 2}x\right) + (21)(1)\left({1 \over 4}x^2\right) + (35)(1)\left(-{1 \over 8}x^3\right) + ... \\
& = 1 - {7 \over 2}x + {21 \over 4}x^2 - {35 \over 8}x^3 + ...
\end{align}
4(ii) Hence, estimate the value of $(0.999)^7$, correct to $5$ decimal places.
Answer: $ 0.993 \phantom{.} 02 $
Solutions
\begin{align}
\text{Let } 1 - {1 \over 2}x & = 0.999 \\
-{1 \over 2}x & = 0.999 - 1 \\
-{1 \over 2}x & = -0.001 \\
{1 \over 2}x & = 0.001 \\
x & = 0.002 \\
\\ \\
\text{From (i), }
\left(1 - {1 \over 2}x\right)^7 & = 1 - {7 \over 2}x + {21 \over 4}x^2 - {35 \over 8}x^3 + ... \\
\\
\text{Let } & x = 0.002, \\
\left[ 1 - {1 \over 2}(0.002) \right]^7
& = 1 - {7 \over 2}(0.002) + {21 \over 4}(0.002)^2 - {35 \over 8}(0.002)^3 + ... \\
(0.999)^7 & = 0.993 \phantom{.} 020 \\
& \approx 0.993 \phantom{.} 02 \phantom{0} \text{ (5 d.p.)}
\end{align}
Binomial expansion to the power of n
5(i) Write down the first three terms in the expansion of $ \left(3 - {x \over 9}\right)^n $, where $n$ is a positive integer greater than $2$, in ascending powers of $x$.
(from Think A Maths Workbook Worksheet 5B)
Answer: $ 3^n - n 3^{n - 3} x + {n(n - 1) 3^{n - 6} \over 2} x^2 + ... $
Solutions
\begin{align}
\left(3 - {x \over 9}\right)^n & = 3^n + {n \choose 1} (3)^{n - 1} \left(-{x \over 9}\right)
+ {n \choose 2} (3)^{n - 2} \left(-{x \over 9}\right)^2 + ... \\
& = 3^n + (n) (3^{n - 1}) \left(-{x \over 9}\right) + {n(n - 1) \over 2} (3^{n - 2})\left(x^2 \over 81\right) + ... \\
& = 3^n - {n 3^{n - 1} \over 9} x + {n(n - 1) 3^{n - 2} \over 2(81)} x^2 + ... \\
& = 3^n - {n 3^{n - 1} \over 3^2} x + {n(n - 1) 3^{n - 2} \over 2(3^4)} x^2 + ... \\
& = 3^n - n 3^{n - 1 -2} x + {n(n - 1)3^{n - 2 - 4} \over 2} x^2 + ... \\
& = 3^n - n 3^{n - 3} x + {n(n - 1) 3^{n - 6} \over 2} x^2 + ...
\end{align}
The first two non-zero terms in the expansion of $(3 + x)\left(3 - {x \over 9}\right)^n$ in ascending powers of $x$ are $a + bx^2$, where $a$ and $b$ are constants.
5(ii) Find the value of $n$.
Hint: Since the term in $x$ does not exist, the coefficient of $x$ is 0.
Answer: $ n = 9 $
Solutions
\begin{align}
(3 + x) \left(3 - {x \over 9}\right)^n
& = (3 + x) \left[ 3^n - n 3^{n - 3} x + {n(n - 1) 3^{n - 6} \over 2} x^2 + ...\right] \\
& = (3)(3^n) - (3)(n 3^{n - 3} x) + (3) \left[{n(n - 1) 3^{n - 6} \over 2} x^2\right]
+ (x)(3^n) + (x)(- n 3^{n - 3} x) + ... \\
& = 3^{n + 1} - n 3^{n - 3 + 1} x + {n(n - 1) 3^{n - 6 + 1} \over 2} x^2 + 3^n x - n 3^{n - 3}x^2 + ... \\
& = 3^{n + 1} - n 3^{n - 2} x + {n(n - 1) 3^{n - 5} \over 2} x^2 + 3^n x - n3^{n - 3}x^2 + ... \\
& = 3^{n + 1} + (-n 3^{n - 2} + 3^n) x + \left[ {n(n - 1) 3^{n - 5} \over 2} - n3^{n - 3}\right]x^2 + ... \\
\\
\text{Comparing } & \text{coefficients of } x, \\
0 & = -n 3^{n - 2} + 3^n \\
n 3^{n - 2} & = 3^n \\
n & = {3^n \over 3^{n - 2}} \\
n & = 3^{n - (n - 2)} \\
n & = 3^2 \\
n & = 9
\end{align}
5(iii) Hence, find the value of $a$ and of $b$.
Answer: $ a = 59 \phantom{.} 049, b = -3645 $
Solutions
\begin{align}
(3 + x) \left(3 - {x \over 9}\right)^n
& = 3^{9 + 1} + [-(9) 3^{9 - 2} + 3^9] x + \left[ {9(9 - 1) 3^{9 - 5} \over 2} - (9)3^{9 - 3}\right]x^2 + ... \\
& = 59 \phantom{.} 049 + (0)x - 3645x^2 + ... \\
\\ \\
\therefore a & = 59 \phantom{.} 049, b = -3645
\end{align}
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| 2008 P1 Question 11 |
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