Sec 3 A Math: Binomial Theorem, practice questions & worked solutions

General term of expansion $\left(2 - \dfrac{1}{2}x\right)^8$ is $$\begin{aligned} T_{r+1} &= \binom{8}{r}(2)^{8-r}\left(-\frac{x}{2}\right)^r\\ &= \binom{8}{r}(-1)^r (2)^{8-2r}\, x^r \end{aligned}$$ $$\begin{aligned} T_3 &= \binom{8}{2}(-1)^2(2)^{8-2(2)}\, x^2\\ &= 448x^2\\ T_4 &= \binom{8}{3}(-1)^3(2)^{8-2(3)}\, x^3\\ &= -224x^3 \end{aligned}$$ $$\begin{aligned} (10 + kx)\left(2 - \frac{1}{2}x\right)^8 &= (10 + kx)(\ldots 448x^2 - 224x^3 \ldots) \end{aligned}$$ $$\begin{aligned} \text{Coeff.\ of } x^3 &= 0\\ (10)(-224) + k(448) &= 0\\ 448k &= 2240\\ k &= 5 \end{aligned}$$

(a)

$$\begin{aligned} \left(\frac{a}{x}+3x^2\right)^6 &= \left(\frac{a}{x}\right)^6+\binom{6}{1}\!\left(\frac{a}{x}\right)^5(3x^2)+\binom{6}{2}\!\left(\frac{a}{x}\right)^4(3x^2)^2+\binom{6}{3}\!\left(\frac{a}{x}\right)^3(3x^2)^3+\cdots\\ &= \frac{a^6}{x^6}+\frac{18a^5}{x^3}+135a^4+540a^3x^3+\cdots \end{aligned}$$

(b)

$$(2x^3-1)\!\left(\frac{a^6}{x^6}+\frac{18a^5}{x^3}+135a^4+540a^3x^3+\cdots\right)$$ since there is no $x^3$ term, $$\begin{aligned} 2(135a^4)-540a^3 &= 0\\ 270a^4-540a^3 &= 0\\ 270a^3(a-2) &= 0\\ a=0\ (\text{rej}),\ &a=2 \end{aligned}$$

$$(1+ax)^n = 1+\binom{n}{1}(ax)+\binom{n}{2}(ax)^2+\cdots = 1+nax+\binom{n}{2}a^2x^2+\cdots$$ By comparing the coefficients of: $$\begin{aligned} x: \binom{n}{1}a &= -3\\ na &= -3\\ a &= -\frac{3}{n} \text{(1)}\\ x^2: \binom{n}{2}a^2 &= \frac{33}{8}\\ \frac{n(n-1)}{2}a^2 &= \frac{33}{8} \text{(2)} \end{aligned}$$ Sub. (1) into (2): $$\begin{aligned} \frac{n(n-1)}{2}\!\left(-\frac{3}{n}\right)^2 &= \frac{33}{8}\\ \frac{9n^2-9n}{2n^2} &= \frac{33}{8}\\ 72n^2-72n &= 66n^2\\ 6n^2-72n &= 0\\ n^2-12n &= 0\\ n(n-12) &= 0\\ n=0\ (\text{rej, since } n\neq 0) \;&\text{or}\; n-12=0\\ n &= 12 \end{aligned}$$ Sub. $n=12$ into (1): $$a=-\frac{3}{12}=-\frac{1}{4}$$ $\therefore\ a=-\dfrac{1}{4}$ and $n=12$.

$$\begin{aligned} (1 + px)^n &= 1 + \binom{n}{1}(px) + \binom{n}{2}(px)^2 + \binom{n}{3}(px)^3 + \cdots\\ &= 1 + npx + \frac{n(n - 1)}{2}p^2x^2 + \frac{n(n - 1)(n - 2)}{6}p^3x^3 + \cdots\\ &= 1 - 12x + 28p^2x^2 + qx^3 + \cdots \end{aligned}$$ By comparing coefficients of $x^2$, $$\begin{aligned} \frac{n(n - 1)}{2} &= 28\\ n^2 - n &= 56\\ n^2 - n - 56 &= 0\\ (n - 8)(n + 7) &= 0\\ n = 8 \ &\text{or} \ n = -7 \ (\text{N.A.}) \end{aligned}$$ By comparing coefficients of $x$, $$\begin{aligned} np &= -12\\ p &= \frac{-12}{8}\\ p &= -1\tfrac{1}{2} \end{aligned}$$ $$\begin{aligned} q &= \frac{8(7)(6)}{6}\left(-1\tfrac{1}{2}\right)^3\\ q &= -189 \end{aligned}$$

(a)

$$(2 + x)^n = 2^n + \binom{n}{1}2^{n-1}x + \binom{n}{2}2^{n-2}x^2 + \dots$$ $$\begin{aligned} \binom{n}{2}2^{n-2} &= 2\binom{n}{1}2^{n-1}\\ \frac{n(n-1)}{2}\left(2^{n-2}\right) &= 2n\left(2^{n-1}\right)\\ \frac{n(n-1)}{2}\left(2^n \times 2^{-2}\right) &= 2n\left(2^n \times 2^{-1}\right) (n \neq 0)\\ \frac{n-1}{2(4)} &= 1\\ n - 1 &= 8\\ n &= 9 \end{aligned}$$

(b)

general term $$\begin{aligned} &= \binom{15}{r}(2x)^{15-r}\left(-\frac{1}{4x^4}\right)^r\\ &= \binom{15}{r}2^{15-r}x^{15-r}\left(-\frac{1}{4}\right)^r x^{-4r}\\ &= \binom{15}{r}2^{15-r}\left(-\frac{1}{4}\right)^r x^{15-5r} \end{aligned}$$ $$\begin{aligned} 15 - 5r &= 0\\ r &= 3 \end{aligned}$$ $$\text{value} = \binom{15}{3}2^{12}\left(-\frac{1}{4}\right)^3 = -29120$$

(a)

$$\begin{aligned} \left(2 - \frac{x}{4}\right)^5 &= (2)^5 + \binom{5}{1}(2)^4\left(-\frac{x}{4}\right) + \binom{5}{2}(2)^3\left(-\frac{x}{4}\right)^2 + \cdots\\ &= 32 - 20x + 5x^2 + \cdots \end{aligned}$$

(b)

$$\begin{aligned} \left(4 + kx + x^2\right)\left(2 - \frac{x}{4}\right)^5 &= \left(4 + kx + x^2\right)\left(32 - 20x + 5x^2 + \cdots\right) \end{aligned}$$ Coeff of $x = -80 + 32k$

Coeff of $x^2 = 52 - 20k$ $$\begin{aligned} -80 + 32k - 20k + 52 &= -4\\ -28 + 12k &= -4\\ 12k &= 24\\ k &= 2 \end{aligned}$$

(a)

General term of $\left(1 - \dfrac{x}{5}\right)^8$ $$\begin{aligned} &= \binom{8}{r}(1)^{8 - r}\!\left(-\frac{x}{5}\right)^r\\ &= \binom{8}{r}\!\left(-\frac{x}{5}\right)^r \end{aligned}$$ $$\begin{aligned} &(7 + 2x)\!\left(1 - \frac{x}{5}\right)^8\\ &= (7 + 2x)\!\left[\ldots + \binom{8}{3}\!\left(-\frac{x}{5}\right)^3 + \binom{8}{2}\!\left(-\frac{x}{5}\right)^2 + \ldots\right] \end{aligned}$$ $$\begin{aligned} x^3 \text{ term} &= 7\binom{8}{3}\!\left(-\frac{x}{5}\right)^3 + 2x\binom{8}{2}\!\left(-\frac{x}{5}\right)^2\\ &= 7(56)\!\left(-\frac{x^3}{125}\right) + 2x(28)\!\left(\frac{x^2}{25}\right)\\ &= -\frac{392}{125}x^3 + \frac{56}{25}x^3 \end{aligned}$$ Coefficient of $x^3 = -\dfrac{112}{125}$

(b)

$$\begin{aligned} \binom{n}{2}3^{n - 2}\!\left(\frac{1}{2}x\right)^2 &= \binom{n}{2}3^{n - 2}\frac{1}{4}x^2\\ \binom{n}{3}3^{n - 3}\!\left(\frac{1}{2}x\right)^3 &= \binom{n}{3}3^{n - 3}\frac{1}{8}x^3 \end{aligned}$$ $$\begin{aligned} \binom{n}{2}3^{n - 2}\frac{1}{4} &= 3 \times \binom{n}{3}3^{n - 3}\frac{1}{8}\\ \binom{n}{2}3^{n - 2}\cdot 2 &= \binom{n}{3}3^{n - 2}\\ \binom{n}{2}\cdot 2 &= \binom{n}{3}\\ \frac{n(n - 1)}{2}\cdot 2 &= \frac{n(n - 1)(n - 2)}{2 \times 3}\\ 6 &= n - 2\\ n &= 8 \end{aligned}$$

(a)

$$(1 + kx)^n = 1 + \binom{n}{1}(kx) + \binom{n}{2}(kx)^2 + \dots$$ $$\binom{n}{1}k = \frac{7}{3} \qquad\qquad \binom{n}{2}k^2 = \frac{7}{3}$$ $$nk = \frac{7}{3} \qquad\qquad \frac{n(n - 1)}{2}k^2 = \frac{7}{3}$$ $$k = \frac{7}{3n}$$ $$\begin{aligned} \frac{n(n - 1)}{2}\left(\frac{7}{3n}\right)^2 &= \frac{7}{3}\\ \frac{n(n - 1)}{2}\left(\frac{7}{3n^2}\right) &= 1\\ \frac{7(n - 1)}{6n} &= 1\\ 7n - 7 &= 6n\\ n &= 7\\ k = \frac{7}{3(7)} &= \frac{1}{3} \end{aligned}$$

(b)

$$\begin{aligned} \text{General Term} &= \binom{12}{r}\left(px^2\right)^{12 - r}\left(-\frac{1}{x^4}\right)^r\\ &= \binom{12}{r}p^{12 - r}x^{24 - 2r}(-1)^r\left(x^{-4}\right)^r\\ &= \binom{12}{r}p^{12 - r}(-1)^r x^{24 - 6r} \end{aligned}$$ Power of $x$: $24 - 6r = 2(12 - 3r)$

Therefore all powers of $x$ is $2(12 - 3r)$, all terms contain only even powers of $x$.