Useful indices laws for simplification:
$ a^m \times a^n = (a^m)(a^n) = \phantom{.} $ $ a^{m + n} $
$ a^m \times b^m = (a^m)(b^m) = \phantom{.} $ $ (ab)^m $
$ (a^m)^n = \phantom{.} $ $ a^{mn} $
$ a^{-n} = \phantom{.} $ $ {1 \over a^n} $
Common questions
1(i) Find the power of $x$ in the general term of the expansion of $\left(2x + {3 \over x}\right)^{12}$.
Answer: $ 12 - 2r $
Solutions
\begin{align*}
T_{r + 1} & = {12 \choose r} (2x)^{12 - r} \left(3 \over x\right)^r \\
& = {12 \choose r} (2)^{12 - r} (x)^{12 - r} (3)^r \left(1 \over x\right)^r
\phantom{0000000} [ (a \times b)^m = a^m \times b^m] \\
& = {12 \choose r} (2^{12 - r}) (x^{12 - r}) (3^r) (x^{-1})^r
\phantom{0000000} \left[ a^{-n} = {1 \over a^n} \right] \\
& = {12 \choose r} (2^{12 - r}) (x^{12 - r}) (3^r) (x^{-r})
\phantom{00000000} [ (a^m)^n = a^{mn}] \\
& = {12 \choose r} (2^{12 - r}) (3^r) (x^{12 - r + (-r)})
\phantom{000000000} [a^m \times a^n = a^{m + n}] \\
& = {12 \choose r} (2^{12 - r}) (3^r) (x^{12 - 2r}) \\
\\
\text{Power of } & x = 12 - 2r
\end{align*}
Hence, find
1(ii) the coefficient of the term in ${1 \over x^8}$,
Answer: $ 15 \phantom{.} 588 \phantom{.} 936 $
Solutions
\begin{align}
{1 \over x^8} & = x^{-8} \\
\\
\text{Let } 12 - 2r & = -8 \\
-2r & = -20 \\
r & = {-20 \over -2} \\
r & = 10 \\
\\
T_{10 + 1} & = {12 \choose 10} (2^{12 - 10}) (3^{10}) (x^{12 - 2(10)}) \\
& = (66)(2^2)(3^{10})(x^{-8}) \\
& = 15 \phantom{.} 588 \phantom{.} 936 x^{-8} \\
& = 15 \phantom{.} 588 \phantom{.} 936 \left(1 \over x^8\right) \\
\\
\text{Coefficient} & = 15 \phantom{.} 588 \phantom{.} 936
\end{align}
1(ii) and the term independent of $x$.
Hint: The term independent of $x$ is the term without $ x $, i.e. the term with $x^0$.
Answer: $ 43 \phantom{.} 110 \phantom{.} 144 $
Solutions
\begin{align}
\text{Let } 12 - 2r & = 0 \\
-2r & = -12 \\
r & = {-12 \over -2} \\
r & = 6 \\
\\
T_{6 + 1} & = {12 \choose 6} (2^{12 - 6}) (3^{6}) (x^{12 - 2(6)}) \\
& = (924)(2^6)(3^6)(x^0) \\
& = 43 \phantom{.} 110 \phantom{.} 144
\end{align}
2. Explain why the binomial expansion of $ \left(x - {1 \over 2x^8}\right)^{15} $ does not have a term that is independent of $x$.
(from A Maths 360 2nd edition Ex 6.2)
Solutions
\begin{align}
T_{r + 1} & = {15 \choose r} (x)^{15 - r} \left(-{1 \over 2x^8}\right)^{r} \\
& = {15 \choose r} (x^{15 - r}) \left(-{1 \over 2}\right)^r \left(1 \over x^8\right)^r \\
& = {15 \choose r} (x^{15 - r}) \left(-{1 \over 2}\right)^r (x^{-8})^r \\
& = {15 \choose r} (x^{15 - r}) \left(-{1 \over 2}\right)^r (x^{-8r}) \\
& = {15 \choose r} \left(-{1 \over 2}\right)^r (x^{15 - r + (-8r)}) \\
& = {15 \choose r} \left(-{1 \over 2}\right)^r (x^{15- 9r}) \\
\\
\text{Let } 15- 9r & = 0 \\
-9r & = -15 \\
r & = {-15 \over -9} \\
r & = {5 \over 3} \\
\\
\text{Since } r \text{ is not} & \text{ an integer, there is no term that is independent of } x
\end{align}
Finding specific ordered term
$ \text{In the expansion of } (a + b)^n, \text{ there are } $$ n + 1 $ $ \text{ terms} $
3. Find the ninth term in the expansion of $ \left(2x - {1 \over x}\right)^{10}$.
Answer: $ 180 x^{-6} $
Solutions
\begin{align}
T_{r + 1} & = {10 \choose r} (2x)^{10 - r} \left(-{1 \over x}\right)^r \\
\\
T_{8 + 1} & = {10 \choose 8} (2x)^{10 - 8} \left(-{1 \over x}\right)^8 \\
T_{9} & = (45)(2x)^2 \left(1 \over x^8\right) \\
& = (45)(4x^2)\left(1 \over x^8\right) \\
& = {180x^2 \over x^8} \\
& = {180 \over x^{6}} \\
& = 180 x^{-6}
\end{align}
4. Find the middle term in the expansion of $\left(x - {2 \over x^2}\right)^{8}$.
Hint: There are $ 9 $ terms in the expansion of $\left(x - {2 \over x^2}\right)^{8}$.
Answer: $ 1120 x^{-4} $
Solutions
\begin{align}
T_{r + 1} & = {8 \choose r} (x)^{8 - r} \left(-{2 \over x^2}\right)^r \\
\\
T_{4 + 1} & = {8 \choose 4} (x)^{8 - 4} \left(-{2 \over x^2}\right)^4
\phantom{000000} [9 \text{ terms in total so } 5 \text{th term is in the middle}] \\
T_5 & = (70)(x^4)\left(16 \over x^8\right) \\
& = {1120x^4 \over x^8} \\
& = {1120 \over x^4} \\
& = 1120 x^{-4}
\end{align}