Binomial Expansion: past paper questions & worked solutions

In $\left(\dfrac38+4x\right)^{12}$, the term in $x^7$ comes from choosing $(4x)^7$ and $\left(\dfrac38\right)^5$.

The coefficient is $\binom{12}{7}\left(\dfrac38\right)^5 4^7$.

$4^7=2^{14}$ and $8^5=2^{15}$, so $\left(\dfrac38\right)^5 4^7=\dfrac{3^5}{2}$.

$\binom{12}{7}=792$, so the coefficient is $792\cdot\dfrac{243}{2}=96228$.

$(2+\dfrac38x)^{10}=2^{10}+\binom{10}{1}2^9\left(\dfrac38x\right)+\binom{10}{2}2^8\left(\dfrac38x\right)^2+\binom{10}{3}2^7\left(\dfrac38x\right)^3+\cdots$.

$2^{10}=1024$ and $\binom{10}{1}2^9\left(\dfrac38\right)x=1920x$.

$\binom{10}{2}2^8\left(\dfrac38\right)^2x^2=1620x^2$.

$\binom{10}{3}2^7\left(\dfrac38\right)^3x^3=810x^3$.

The first four terms are $1024+1920x+1620x^2+810x^3$.

$(2-5x)^8=2^8+8(2^7)(-5x)+\binom82(2^6)(-5x)^2+\binom83(2^5)(-5x)^3+\cdots$.

The first four terms are $256-5120x+44800x^2-224000x^3$.

To use $(2-5x)^8$ for $2.05^8$, set $2-5x=2.05$.

$-5x=0.05$, so $x=-0.01$.

$\left(1-\dfrac16x\right)^9=1+9\left(-\dfrac16x\right)+\binom92\left(-\dfrac16x\right)^2+\binom93\left(-\dfrac16x\right)^3+\cdots$.

The first four terms are $1-\dfrac32x+x^2-\dfrac{7}{18}x^3$.

In $(10x+3)\left(1-\dfrac16x\right)^9$, the $x^3$ coefficient comes from $3$ times the $x^3$ coefficient and $10x$ times the $x^2$ coefficient.

$3\left(-\dfrac{7}{18}\right)+10(1)=-\dfrac76+10=\dfrac{53}{6}$.

For $(3+kx)^7$, the first four terms are $3^7+\binom71 3^6(kx)+\binom72 3^5(kx)^2+\binom73 3^4(kx)^3$.

This gives $2187+5103kx+5103k^2x^2+2835k^3x^3$.

The coefficient of $x^2$ is $5103k^2$ and the coefficient of $x^3$ is $2835k^3$.

Since $5103k^2=3(2835k^3)$ and $k\ne0$, divide by $k^2$ to get $5103=8505k$, hence $k=\dfrac{5103}{8505}=\dfrac35$.

$(1+px)^{10}=1+10px+\binom{10}{2}p^2x^2+\binom{10}{3}p^3x^3+\cdots$.

Since the coefficient of $x$ is $15$, $10p=15$, so $p=\dfrac32$.

$q=\binom{10}{2}p^2=45\left(\dfrac32\right)^2=\dfrac{405}{4}$.

$r=\binom{10}{3}p^3=120\left(\dfrac32\right)^3=405$.

$(1-4x)^7=1+7(-4x)+\binom72(-4x)^2+\binom73(-4x)^3+\cdots$.

The first four terms are $1-28x+336x^2-2240x^3$.

In $(5+kx)(1-4x)^7$, the $x^2$ coefficient comes from $5(336x^2)$ and $(kx)(-28x)$.

$5(336)-28k=1316$, so $1680-28k=1316$.

$28k=364$, hence $k=13$.

In $(1+kx)^{16}$, the coefficient of $x$ is $16k$.

Since the coefficient of $x$ is $-4$, $16k=-4$, so $k=-\dfrac14$.

The coefficient of $x^2$ is $\binom{16}{2}k^2=120\left(\dfrac1{16}\right)=\dfrac{15}{2}$.

For $k=-\dfrac14$, the $x^2$ coefficient in $(1+kx)^{16}$ is $\dfrac{15}{2}$ and the $x^3$ coefficient is $\binom{16}{3}\left(-\dfrac14\right)^3=-\dfrac{35}{4}$.

In $g(x)=\left(2+\dfrac{16}{x}\right)(1+kx)^{16}$, the $x^2$ term comes from $2\left(\dfrac{15}{2}x^2\right)$ and $\dfrac{16}{x}\left(-\dfrac{35}{4}x^3\right)$.

The required term is $15x^2-140x^2=-125x^2$.

$(1+\dfrac{x}{4})^{12}=1+\binom{12}{1}\dfrac{x}{4}+\binom{12}{2}\left(\dfrac{x}{4}\right)^2+\binom{12}{3}\left(\dfrac{x}{4}\right)^3+\cdots$.

This gives $1+3x+\dfrac{33}{8}x^2+\dfrac{55}{16}x^3$.

$\dfrac{x^2+8}{x^5}=x^{-3}+8x^{-5}$.

The $x^{-3}$ part needs the $x^3$ coefficient, which is $\dfrac{55}{16}$. The $8x^{-5}$ part needs the $x^5$ coefficient, which is $8\binom{12}{5}\left(\dfrac14\right)^5=8\cdot\dfrac{99}{128}=\dfrac{99}{16}$.

The term independent of $x$ is $\dfrac{55}{16}+\dfrac{99}{16}=\dfrac{154}{16}=\dfrac{77}{8}$.

$(2+ax)^6=2^6+6(2^5)(ax)+\binom62(2^4)(ax)^2+\cdots$.

The first three terms are $64+192ax+240a^2x^2$.

$\left(3+\dfrac1x\right)^2=9+6x^{-1}+x^{-2}$.

The constant term in $\left(3+\dfrac1x\right)^2(2+ax)^6$ is $9(64)+6(192a)+240a^2$.

$576+1152a+240a^2=576$, so $240a^2+1152a=0$.

$48a(5a+24)=0$. Since $a\ne0$, $5a+24=0$, so $a=-\dfrac{24}{5}$.

$(3+2x)^6=3^6+\binom61 3^5(2x)+\binom62 3^4(2x)^2+\binom63 3^3(2x)^3+\cdots$.

This gives $729+2916x+4860x^2+4320x^3$.

In $\left(2x^2-\dfrac{1}{6x}\right)(3+2x)^6$, the $x^2$ term comes from $2x^2$ times the constant term and $-\dfrac{1}{6x}$ times the $x^3$ term.

The coefficient is $2(729)-\dfrac16(4320)=1458-720=738$.

In $(3+px)^5$, the constant term is $3^5=243$, so $A=243$.

The coefficient of $x$ is $5(3^4)p=405p$, so $B=405p$.

The coefficient of $x^3$ is $\binom53(3^2)p^3=10\cdot9p^3=90p^3$, so $D=90p^3$.

$405p=18(90p^3)=1620p^3$.

Since $p<0$, $p\ne0$. Divide by $405p$ to get $1=4p^2$, so $p=\pm\dfrac12$. Therefore $p=-\dfrac12$.

The coefficient of $x^2$ is $\binom52(3^3)p^2=10\cdot27\cdot\dfrac14=\dfrac{135}{2}$.

$\left(2+\dfrac{kx}{8}\right)^7$ begins with $2^7+7(2^6)\left(\dfrac{kx}{8}\right)+\binom72(2^5)\left(\dfrac{kx}{8}\right)^2+\binom73(2^4)\left(\dfrac{kx}{8}\right)^3$.

This gives $128+56kx+\dfrac{21}{2}k^2x^2+\dfrac{35}{32}k^3x^3$.

The coefficients of $x$, $x^2$ and $x^3$ are $56k$, $\dfrac{21}{2}k^2$ and $\dfrac{35}{32}k^3$.

For three terms in an arithmetic progression, twice the middle term equals the sum of the outer terms: $21k^2=56k+\dfrac{35}{32}k^3$.

Since $k\ne0$, divide by $k$ and multiply by $32$: $672k=1792+35k^2$.

So $35k^2-672k+1792=0$. The discriminant is $672^2-4(35)(1792)=200704=448^2$.

$k=\dfrac{672\pm448}{70}$, giving $k=16$ or $k=\dfrac{16}{5}$.

Dividing by $4x-5$ means the remainder is $f\left(\dfrac54\right)$.

$f\left(\dfrac54\right)=\left(2-\dfrac{5k}{4}\right)^5=\dfrac{243}{32}$.

Since $\dfrac{243}{32}=\left(\dfrac32\right)^5$, we have $2-\dfrac{5k}{4}=\dfrac32$.

$\dfrac{5k}{4}=\dfrac12$, so $k=\dfrac25$.

With $k=\dfrac25$, $f(x)=\left(2-\dfrac25x\right)^5$.

$2^5+5(2^4)\left(-\dfrac25x\right)+10(2^3)\left(-\dfrac25x\right)^2=32-32x+\dfrac{64}{5}x^2$.

For a polynomial, the gradient at $x=0$ is the coefficient of $x$ in the expansion. Therefore $f'(0)=-32$.

$(2-\dfrac{kx}{4})^8=2^8+\binom81 2^7\left(-\dfrac{kx}{4}\right)+\binom82 2^6\left(-\dfrac{kx}{4}\right)^2+\binom83 2^5\left(-\dfrac{kx}{4}\right)^3+\cdots$.

The first four terms are $256-256kx+112k^2x^2-28k^3x^3$.

For $f(x)=(5-3x)(2-\dfrac{kx}{4})^8$, the constant term is $5(256)=1280$.

The coefficient of $x$ is $5(-256k)-3(256)=-1280k-768$.

The constant term is $3$ times the coefficient of $x$, so $1280=3(-1280k-768)$.

$1280=-3840k-2304$, hence $k=-\dfrac{3584}{3840}=-\dfrac{14}{15}$.

In $(3+ax)^6$, the $x^4$ term is $\binom64 3^2(ax)^4$.

$\binom64 3^2a^4x^4=15\cdot9a^4x^4=135a^4x^4$. Since this term is $540x^4$, $135a^4=540$.

$a^4=4$, so the real possible values are $a=\sqrt2$ or $a=-\sqrt2$.

The constant term from $\dfrac1{81}(3+ax)^6$ is $\dfrac1{81}\cdot3^6=9$.

The $x^6$ term in $(3+ax)^6$ is $a^6x^6$. Since $a=\pm\sqrt2$, $a^6=8$, so $\dfrac1{x^6}\cdot a^6x^6$ contributes $8$.

The term independent of $x$ is $9+8=17$.

$(2+\dfrac{x}{8})^{13}=2^{13}+\binom{13}{1}2^{12}\dfrac{x}{8}+\binom{13}{2}2^{11}\left(\dfrac{x}{8}\right)^2+\binom{13}{3}2^{10}\left(\dfrac{x}{8}\right)^3+\cdots$.

This gives $8192+6656x+2496x^2+572x^3+\cdots$.

Since $2.0125=2+0.0125=2+\dfrac{0.1}{8}$, use $x=0.1$ in the expansion.

$8192+6656(0.1)+2496(0.1)^2+572(0.1)^3=8883.132$.

The next terms in the expansion are positive because $x=0.1>0$, so stopping after the $x^3$ term gives an underestimate.

$(2+px)^6=2^6+\binom61 2^5(px)+\binom62 2^4(px)^2+\cdots$.

The first three terms are $64+192px+240p^2x^2$.

In $(3-\dfrac12x)(2+px)^6$, the $x^2$ coefficient is $3(240p^2)-\dfrac12(192p)$.

So $720p^2-96p=-\dfrac34$.

Multiplying by $4$ gives $2880p^2-384p+3=0$, or $960p^2-128p+1=0$.

$p=\dfrac{128\pm112}{1920}$, so $p=\dfrac18$ or $p=\dfrac1{120}$.

The 2nd term of $(3+2x)^6$ is $\binom61 3^5(2x)=6\cdot243\cdot2x=2916x$.

The 3rd term is $\binom62 3^4(2x)^2=15\cdot81\cdot4x^2=4860x^2$.

The 5th term is $\binom64 3^2(2x)^4=15\cdot9\cdot16x^4=2160x^4$.

For three consecutive terms in a geometric series, the middle term squared equals the product of the two outer terms.

$(4860x^2)^2=(2916x)(2160x^4)$.

For the non-zero geometric case, divide by $x^4$: $4860^2=2916\cdot2160\,x$, so $x=\dfrac{4860^2}{2916\cdot2160}=\dfrac{15}{4}$.

In the different geometric series, first term $=\sin^2\theta$, common ratio $=2\cos\theta$, and $S_\infty=\dfrac85$.

$\dfrac{\sin^2\theta}{1-2\cos\theta}=\dfrac85$.

$5\sin^2\theta=8(1-2\cos\theta)$. With $\sin^2\theta=1-\cos^2\theta$, this becomes $5(1-\cos^2\theta)=8-16\cos\theta$.

Rearranging gives $5\cos^2\theta-16\cos\theta+3=0$.

$5\cos^2\theta-16\cos\theta+3=0$ gives $\cos\theta=3$ or $\cos\theta=\dfrac15$. The possible value is $\cos\theta=\dfrac15$.

The 2nd term is first term times common ratio: $\sin^2\theta(2\cos\theta)=\left(1-\dfrac{1}{25}\right)\cdot\dfrac25=\dfrac{24}{25}\cdot\dfrac25=\dfrac{48}{125}$.