Sec 4 A Math: Binomial Theorem, practice questions & worked solutions

(a)

$$\left(px-\frac{q}{x}\right)^{\!n}$$ $$\begin{aligned} T_{r+1} &= \binom{n}{r}(px)^{n-r}\!\left(-\frac{q}{x}\right)^{\!r}\\ &= \binom{n}{r}\!\left(p^{\,n-r}\right)(-q)^r\,x^{\,n-2r} \end{aligned}$$

(b)

5th term $\Rightarrow r=4$. $$\begin{aligned} T_5 &= \binom{n}{4}\!\left(p^{\,n-4}\right)(-q)^4\,x^{\,n-8}\\ n-8 &= 0\\ n &= 8 \end{aligned}$$

(c)

$$\begin{aligned} \binom{8}{4}\!\left(p^4\right)(-q)^4 &= 1120\\ 70p^4q^4 &= 1120\\ (pq)^4 &= 16\\ pq &= 2\\ q &= \frac{2}{p} \end{aligned}$$ Sub $q=\dfrac{2}{p}$ into $p-q=1$: $$\begin{aligned} p-\frac{2}{p} &= 1\\ p^2-2 &= p\\ p^2-p-2 &= 0\\ (p+1)(p-2) &= 0\\ p=-1\ \text{(rej.)} \;&\text{or}\; p=2\\ \therefore\; q &= 1 \end{aligned}$$

(a)

$$\begin{aligned} \left(x+\frac{k}{x^2}\right)^5 &= x^5+\binom{5}{1}x^4\!\left(\frac{k}{x^2}\right)+\binom{5}{2}x^3\!\left(\frac{k}{x^2}\right)^2+\binom{5}{3}x^2\!\left(\frac{k}{x^2}\right)^3+\cdots\\ &= x^5+5kx^2+\frac{10k^2}{x}+\frac{10k^3}{x^4}+\cdots \end{aligned}$$

(b)

$$\left(5x+\frac{3}{x^2}\right)\!\left(x+\frac{k}{x^2}\right)^5 = \left(5x+\frac{3}{x^2}\right)\!\left(x^5+5kx^2+\frac{10k^2}{x}+\frac{10k^3}{x^4}+\cdots\right)$$ constant term $=5$ $$\begin{aligned} 5\times 10k^2+3\times 5k &= 5\\ 10k^2+3k-1 &= 0\\ (5k-1)(2k+1) &= 0\\ k=\frac{1}{5} \;&\text{or}\; k=-\frac{1}{2}\ (\text{reject}) \end{aligned}$$

(c)

$$\begin{aligned} T_{r+1} &= \binom{n}{r}x^{n-r}\!\left(\frac{k}{x^2}\right)^r\\ &= \binom{n}{r}k^r x^{n-3r} \end{aligned}$$ For term independent of $x$, $n=3r$.

Since $n$ must be a multiple of 3, possible values of $n$ are 6 and 9.

(a)

$$\begin{aligned} \text{General term} &= \binom{8}{r}x^{8-r}\!\left(-\frac{k}{x^3}\right)^r\\ &= \binom{8}{r}x^{8-r}(-1)^r k^r\!\left(x^{-3}\right)^r\\ &= \binom{8}{r}(-1)^r k^r x^{8-4r} \end{aligned}$$ When $8-4r=0$, $r=2$. $$\begin{aligned} \binom{8}{2}(-1)^2 k^2 x^{8-4(2)} &= 7\\ 28k^2 &= 7\\ k^2 &= \frac{1}{4}\\ k &= \frac{1}{2} \end{aligned}$$

(b)

When $8-4r=-4$, $r=3$. $$\begin{aligned} T_4 &= \binom{8}{3}(-1)^3\!\left(\frac{1}{2}\right)^3 x^{-4}\\ &= -7x^{-4}\\ (1+x^4)\!\left(x-\frac{k}{x^3}\right)^8 &= (1+x^4)(\ldots+7-7x^{-4}+\ldots)\\ &= \ldots+7-7+\ldots = \ldots+0+\ldots \end{aligned}$$ There is no constant term in the expansion.

(a)

$$\begin{aligned} (1 + 3x)^8 &= 1 + 8(3x) + \binom{8}{2}(3x)^2 + \cdots\\ &= 1 + 24x + 252x^2 + \cdots \end{aligned}$$

(b)

$$\begin{aligned} \left(2 + 6x + 3x^2\right)^8 &= 2^8\left(1 + 3x + \frac{3}{2}x^2\right)^8\\ &= 256\left(1 + 3\left(x + \frac{1}{2}x^2\right)\right)^8\\ &= 256\left(1 + 24\left(x + \frac{1}{2}x^2\right) + 252\left(x + \frac{1}{2}x^2\right)^2 + \cdots\right)\\ &= 256\left(1 + 24x + 12x^2 + 252x^2 + \cdots\right)\\ &= 256\left(1 + 24x + 264x^2 + \cdots\right)\\ &= 256 + 6144x + 67584x^2 + \cdots \end{aligned}$$

(c)

$$\begin{aligned} \left(2 - \frac{1}{3x}\right)^2 (1 + 3x)^8 &= \left(2 - \frac{1}{3x}\right)^2 \left(1 + 24x + 252x^2 + \cdots\right)\\ &= \left(4 - \frac{4}{3x} + \frac{1}{9x^2}\right)\left(1 + 24x + 252x^2 + \cdots\right) \end{aligned}$$ $$\begin{aligned} \text{constant term} &= 4(1) - \frac{4}{3}(24) + \frac{1}{9}(252)\\ &= 0 \end{aligned}$$ There is no constant term.

(a)

$$\begin{aligned} T_{r+1} &= \binom{6}{r}\left(x^4\right)^{6-r}\left(-\frac{1}{kx^2}\right)^r\\ &= \binom{6}{r}\left(-\frac{1}{k}\right)^r\left(x^{24-4r}\right)x^{-2r}\\ &= \binom{6}{r}\left(-\frac{1}{k}\right)^r x^{24-6r} \end{aligned}$$ Powers of $x = 24 - 6r$

Since 24 and $6r$ are even numbers, subtraction between even numbers will give an even number.

$\therefore$ there are only even powers of $x$ in this expansion.

(b)

$$\begin{aligned} 24 - 6r &= 0\\ r &= 4\\ \binom{6}{4}\left(-\frac{1}{k}\right)^4 &= \frac{5}{27}\\ \left(-\frac{1}{k}\right)^4 &= \frac{1}{81}\\ k &= 3 \end{aligned}$$

(c)

$$\begin{aligned} 24 - 6r &= -6\\ r &= 5 \end{aligned}$$ Coefficient of $x^{-6} = \dbinom{6}{5}\left(-\dfrac{1}{3}\right)^5 = -\dfrac{6}{243} = -\dfrac{2}{81}$

For the expansion $\left(2 - 3x^6\right)\!\left(x^4 - \dfrac{1}{kx^2}\right)^6$ $$\begin{aligned} \text{Term independent of } x &= 2\left(\frac{5}{27}\right) - 3\left(-\frac{6}{243}\right)\\ &= \frac{10}{27} + \frac{2}{27}\\ &= \frac{12}{27}\\ &= \frac{4}{9} \end{aligned}$$

(a)

$$\begin{aligned} \text{General term for } \left(\frac{1}{2x} - ax\right)^5 &= \binom{5}{r}\left(\frac{1}{2x}\right)^{5-r}(-ax)^r\\ &= \binom{5}{r}\left(\frac{1}{2}\right)^{5-r}(-a)^r x^{2r-5} \end{aligned}$$ $$\begin{aligned} 2r - 5 = 1 \ \text{and} \ 2r - 5 &= -1\\ r = 3 \ \text{and} \ r &= 2 \end{aligned}$$ $$\begin{aligned} \text{Term with } x^{-1} &= \binom{5}{2}\left(\frac{1}{2}\right)^{5-2}(-a)^2 x^{4-5}\\ &= \frac{5}{4}a^2\left(\frac{1}{x}\right)\\ \text{Term with } x &= \binom{5}{3}\left(\frac{1}{2}\right)^{5-3}(-a)^3 x^{6-5}\\ &= -\frac{5}{2}a^3 x \end{aligned}$$ $$\begin{aligned} \left(2 + x^2\right)\left(\frac{1}{2x} - ax\right)^5 &= \left(2 + x^2\right)\!\left(-\frac{5}{2}a^3 x + \frac{5}{4}a^2\left(\frac{1}{x}\right) + \cdots\right)\\ &= \left(-5a^3 + \frac{5}{4}a^2\right)x + \cdots \end{aligned}$$ $$\begin{aligned} -5a^3 + \frac{5}{4}a^2 &= 0\\ \frac{5}{4}a^2(-4a + 1) &= 0\\ a = 0 \ (\text{rejected}) \ \text{or} \ a &= \frac{1}{4} \end{aligned}$$

(b)

$$\begin{aligned} \left(2 + \frac{x^2}{2}\right)^5 &= 2^5 + \binom{5}{1}2^4\left(\frac{x^2}{2}\right) + \binom{5}{2}2^3\left(\frac{x^2}{2}\right)^2 + \cdots\\ &= 32 + 40x^2 + 20x^4 + \cdots \text{---- (1)} \end{aligned}$$ Solving $\left(2 + \dfrac{x^2}{2}\right)^5 = 2.005^5$ gives $x = 0.1$

Substitute $x = 0.1$ into (1): $$\begin{aligned} 2.005^5 &= 32 + 40(0.1)^2 + 20(0.1)^4\\ &= 32 + 40(0.01) + 20(0.0001)\\ &= 32.402 \end{aligned}$$

(a)

$\left(ax + \dfrac{2}{x}\right)^6$

Term $$\binom{6}{r}(ax)^{6-r}\left(\frac{2}{x}\right)^r$$ $$\frac{x^{6-r}}{x^r} = x^{6-2r}$$ Powers of $x$ $$6 - 2r = 2(3 - r)$$ Since the power/exponent can be expressed as a multiple of 2, therefore all the expansion only contains even powers of $x$.

(b)

$$\left(1 - \frac{x}{a}\right)^3 = 1 - \frac{3x}{a} + \frac{3x^2}{a^2} + \cdots$$

(c)

$\left(ax + \dfrac{2}{x}\right)^6$

Term in $\dfrac{1}{x^4}$ $$\begin{aligned} x^{6-2r} &= x^{-4}\\ 6 - 2r &= -4\\ r &= 5 \end{aligned}$$ $$\frac{1}{x^4}\text{ term } = \binom{6}{5}(ax)\left(\frac{2}{x}\right)^5 = 192a\left(\frac{1}{x^4}\right)$$ $\left(1 - \dfrac{x}{a}\right)^3$

Term in $x^2 = \dfrac{3}{a^2}(x^2)$ from (b) $$\begin{aligned} 192a &= \frac{3}{a^2} \times 1728\\ a^3 &= \frac{3 \times 1728}{192}\\ a^3 &= 27\\ a &= 3 \end{aligned}$$

(d)

Term independent of $x$ $$\begin{aligned} \left(1 - \frac{3x}{a} + \frac{3x^2}{a^2} + \cdots\right)\!\left(1 + \frac{1}{x}\right) &= (1)(1) + \left(-\frac{3x}{3}\right)\!\left(\frac{1}{x}\right) + \cdots \end{aligned}$$ There is no term independent of $x$.