Question

Using binomial theorem, write down the expansions of the following: $\left(i\right)(2x+3y{)}^5$

$\left(ii\right)(2x-3y{)}^4$

$\left(iii\right){(x-\frac{1}{x})}^6$

$\left(iv\right)(1-3x{)}^7$

$\left(v\right){(ax-\frac{b}{x})}^6$

$\left(vi\right){(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}})}^6$

$\left(vii\right)(\sqrt[3]{3x}-\sqrt[3]{3a}{)}^6$

$\left(viii\right)(1+2x-3x^2{)}^5$

$\left(ix\right){(x+1-\frac{1}{x})}^3$

$\left(x\right)(1-2x+3x^2{)}^3$

Answer 1

$(2x+3y{)}^5$

The expansion of $(x+y{)}^n$ has n+1 term, so the expansion of $(2x+3y{)}^5$ has 6 terms.

Using binomial theorem, we have

$(2x+3y{)}^5{=}^5{C}_0(2x{)}^5\left(3y{)}^0{+}^5{C}_1\right(2x{)}^4\left(3y{)}^1{+}^5{C}_2\right(2x{)}^3\left(3y{)}^2{+}^5{C}_3\right(2x{)}^2\left(3y{)}^3{+}^5{C}_4\right(2x\left)\right(3y{)}^4{}^{}{+}^5{C}_5\left(2x{)}^0\right(3y{)}^5$

$=2^5{x}^5+5\times 2^4\times 3\times x^4\times y+10\times 2^3\times 3^2\times x^3\times y^2+10\times 2^2\times 3^3\times x^2\times y^3+5\times 2\times 3^4\times x\times y^4+3^5{y}^5$

$=32x^5+240x^{4}y+720x^3{y}^2+1080x^2{y}^3+810x{y}^4+243y^5$

$\left(ii\right)(2x-3y{)}^4$

The expansion of $(x+y{)}^n$ has n +1 terms so the expansions $(2x-3y{)}^4$ has 5 terms Using binomial therorem, we have

$(2x-3y{)}^4{=}^4{C}_0(2x{)}^4\left(3y{)}^0{-}^4{C}_1\right(2x{)}^3\left(3y{)}^1{+}^4{C}_2\right(2x{)}^2\left(3y{)}^2{-}^4{C}_3\right(2x{)}^1\left(3y{)}^3{+}^4{C}_4\right(2x{)}^0(3y{)}^4$

$=2^4{x}^4-4\times 2^3\times 3x^{3}y+6\times 2^2\times 3^2\times x^2{y}^2-4\times 2\times 3^3\times x{y}^3+3^4{y}^4$

$=16x^4-96x^{3}y+216x^2{y}^2-216x{y}^3+81y^4$

$\left(iii\right){(x-\frac{1}{x})}^6$

The expansion of $(x+y{)}^n$ has n+1 terms so the expansion of ${(x-\frac{1}{x})}^6$ has 7 terms. Using binomial theorem, we get

${(x-\frac{1}{x})}^6{=}^6{C}_0{x}^6\left(\frac{1}{x}\right){-}^6{C}_1{x}^5\frac{1}{x}{+}^6{C}_2{x}^4{\left(\frac{1}{x}\right)}^2{+}^6{C}_4{x}^2{\left(\frac{1}{x}\right)}^4{-}^6{C}_{5}x{\left(\frac{1}{x}\right)}^5{+}^6{C}_6{x}^0{\left(\frac{1}{x}\right)}^6$

$=x^6{-}^6{x}_4{+}^{15}{x}^2-20+\frac{15}{x^2}-\frac{6}{x^4}+\frac{1}{x^6}$

$\left(iv\right)(1-3x{)}^n$

The expansion of $(x+y{)}^n$ has n+1 terms so the expansion $(1-3x{)}^7$ has 8 terms.

Using bionimal theorem to expand, we get

$(1-3x{)}^7{=}^7{C}_0(1{)}^7\left(3x{)}^0{-}^7{C}_1\right(3x\left){+}^7{C}_2\right(3x{)}^2{-}^7{C}_3\left(3x{)}^3{+}^7{C}_4\right(3x{)}^4{-}^7{C}_5\left(3x{)}^5{-}^7{C}_6\right(3x{)}^6{}^{}{+}^7{C}_7{C}_7(3x{)}^7$

$=1-21x+21\times 9x^2-35\times 3^3{x}^3-21x{3}^5{x}^5+7x{e}^6{x}^6-3^7{x}^7$

$=1-21x+189x^2-945x^3+2835x^4+5103x^6-2187x^7$

$\left(v\right){(ax-\frac{b}{x})}^6$

The expansion of (x+y)n has n+1 terms so the expansion of ${(ax-\frac{b}{x})}^6$ has 7 terms.

Using binomial theorem to expand, we get

${(ax-\frac{b}{x})}^6{=}^6{C}_0\left(ax{)}^6{\left(\frac{b}{x}\right)}^0{-}^6{C}_1\right(ax{)}^5\left(\frac{b}{x}\right){+}^6{C}_2\left(ax{)}^4{\left(\frac{b}{x}\right)}^2{-}^6{C}_3\right(ax{)}^3{\left(\frac{b}{x}\right)}^3{+}^6{C}_4\left(ax{)}^2{\left(\frac{b}{x}\right)}^4{}^{}{-}^6{C}_5\right(ax\left){\left(\frac{b}{x}\right)}^5{+}^6{C}_6\right(ax{)}^0{\left(\frac{b}{x}\right)}^6$

$=a^6{x}^6-6a^5{x}^5+15a^4{x}^4\frac{b^2}{x^2}-20a^3{b}^3+15a^2\frac{b^4}{x^2}-6a\frac{b^5}{x^2}-6a\frac{b^5}{x^4}+\frac{b^6}{x^6}$

$=a^6{x}^6-6a^5{x}^{4}b\frac{b}{x}+15a^4{b}^2{x}^2-20a^3{b}^3+15\frac{a^2{b}^4}{x^2}-6\frac{a{b}^5}{x^4}+\frac{b^6}{x^6}$

$\left(vi\right){(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}})}^6$

The expansion of (x+y)^n has n+1 terms so the expansion of ${(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}})}^6$ has 7 terms.

Using binomial theorem to expand, we get

${(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}})}^6{=}^6{C}_0{\left(\sqrt{\frac{x}{a}}\right)}^6{\left(\sqrt{\frac{a}{x}}\right)}^0{-}^6{C}_1{\left(\sqrt{\frac{x}{a}}\right)}^5{\left(\sqrt{\frac{a}{x}}\right)}^1{-}^6{C}_2{\left(\sqrt{\frac{x}{a}}\right)}^4{\left(\sqrt{\frac{a}{x}}\right)}^2$

${=}^6{C}_3{\left(\sqrt{\frac{x}{a}}\right)}^3{\left(\sqrt{\frac{a}{x}}\right)}^3{+}^6{C}_4{\left(\sqrt{\frac{x}{a}}\right)}^2{\left(\sqrt{\frac{a}{x}}\right)}^4{-}^6{C}_5\left(\sqrt{\frac{x}{a}}\right){\left(\sqrt{\frac{a}{x}}\right)}^5{+}^6{C}_6{\left(\sqrt{\frac{x}{a}}\right)}^0{\left(\sqrt{\frac{a}{x}}\right)}^6$

$={\left(\frac{x}{a}\right)}^{\frac{1}{2}\times 6}-6{\left(\frac{x}{a}\right)}^{\frac{1}{2}\times 5}{\left(\frac{a}{x}\right)}^{\frac{1}{2}}+15{\left(\frac{x}{a}\right)}^{\frac{1}{2}\times 4}{\left(\frac{a}{x}\right)}^{2\times \frac{1}{2}}-20{\left(\frac{x}{a}\right)}^{3\times \frac{1}{2}}{\left(\frac{a}{x}\right)}^{3\times \frac{1}{2}}{\left(\frac{a}{x}\right)}^{3\times \frac{1}{2}}{\left(\frac{a}{x}\right)}^{4\times \frac{1}{2}}-6{\left(\frac{x}{a}\right)}^{\frac{1}{2}}{\left(\frac{a}{x}\right)}^{5\times \frac{1}{2}}+{\left(\frac{a}{x}\right)}^{6\times \frac{1}{2}}$

$=\frac{x^3}{a^3}-6\frac{x^{\frac{5}{2}-\frac{1}{2}}}{a^{\frac{5}{2}-\frac{1}{2}}}+15\times \frac{x^2\times a}{a^2\times x}-20x\frac{x^{\frac{3}{2}-\frac{3}{2}}}{a^{\frac{3}{2}-\frac{3}{2}}}+15x\frac{x}{a}\times \frac{a^2}{x^2}-6x\frac{x^{\frac{1}{2}-\frac{5}{2}}}{a^{\frac{1}{2}-\frac{5}{2}}}+\frac{a^3}{x^3}$

$=\frac{x^3}{a^3}-\frac{6x^2}{a^2}+\frac{15x}{a}-20+\frac{15a}{x}-\frac{6a^2}{x^2}+\frac{a^3}{x^3}$

$\left(vii\right)(\sqrt[3]{3x}-\sqrt[3]{3a}{)}^6$

$(\sqrt[3]{3x}-\sqrt[3]{3a}{)}^6$

$={(}^6{C}_0\left)\right(\sqrt[3]{3x}{)}^6(-\sqrt[3]{3a}{)}^0+{(}^6{C}_1)\left(\sqrt[3]{3x}{)}^5\right(-\sqrt[3]{3a}{)}^1+{(}^6{C}_2\left)\right(\sqrt[3]{3x}{)}^4(-\sqrt[3]{3a}{)}^2$

${(}^6{C}_3\left)\right(\sqrt[3]{3x}{)}^3(-\sqrt[3]{3a}{)}^3+{(}^6{C}_4)\left(\sqrt[3]{3x}{)}^2\right(-\sqrt[3]{3a}{)}^4+{(}^6{C}_5\left)\right(\sqrt[3]{3x}{)}^1(-\sqrt[3]{3a}{)}^5{(}^6{C}_6)\left(\sqrt[3]{3x}{)}^0\right(-\sqrt[3]{3a}{)}^6$

$=x^2-6x^{\frac{5}{3}}{a}^{\frac{1}{3}}+15x^{\frac{4}{3}}{a}^{\frac{2}{3}}-20ax+15x^{\frac{2}{3}}{a}^{\frac{4}{3}}-6x^{\frac{1}{3}}{a}^{\frac{5}{3}}+a^2$

$\left(vii\right)(1+2x-3x^2{)}^5$

Let y =1+2x, then

$(1+2x-3x^2{)}^5=(y-3x^2{)}^5$

The expansion of $(x+y{)}^n$ has n+1 terms so the expansion of $(y-3x^2{)}^5$ has 6 terms.

Using binomial theorem to expand, we get

$(y-3x^2{)}^5{=}^{5}C-0y^5(3x^2{)}^0{-}^5{C}_1{y}^4\left(3x^2{)}^1{+}^5{C}_2{y}^3\right(3x^2{)}^2{-}^5{C}_3{y}^2\left(3x^2{)}^3{+}^5{C}_{4}y\right(3x^2{)}^4{-}^5{C}_5{y}^0(3x^2{)}^5$

$=y^5-5y^{4}3x^2+10y^{3}9x^4-10y^2\left(27x^6\right)+5y81x^8-243x^{10}$

$y^5=(1+2x{)}^5{=}^5{C}_0{+}^5{C}_1(2x{)}^1{+}^5{C}_2\left(2x{)}^2{+}^5{C}_3\right(2x{)}^3{+}^5{C}_4\left(2x{)}^4{+}^5{C}_4\right(2x{)}^4{+}^5{C}_5(2x{)}^5$

$y^4=(1+2x{)}^4{=}^4{C}_0{+}^4{C}_1(2x{)}^1{+}^4{C}_2\left(2x{)}^2{+}^4{C}_3\right(2x{)}^3{+}^4{C}_4(2x{)}^4$

$y^3=(1+2x{)}^3{=}^3{C}_0{+}^3{C}_1(2x\left){+}^3{C}_2\right(2x{)}^2{+}^3{C}_3(2x{)}^3$

$y^2=(1+2x{)}^2{=}^2{C}_0{+}^2{C}_1(2x\left){+}^2{C}_2\right(2x{)}^2$

y =(1+2x)

Substituting the values of powers of y in the equation above, we get,

$(1+2x-3x^2{)}^5={[}^5{C}_0{+}^5{C}_1(2x{)}^1{+}^5{C}_2\left(2x{)}^2{+}^5{C}_3\right(2x{)}^3{+}^5{C}_4\left(2x{)}^4{+}^5{C}_5\right(2x{)}^5]$

$=15x^2{[}^4{C}_0{+}^4{C}_1\left(2x{)}^1{+}^4{C}_2\right(2x{)}^2{+}^4{C}_3\left(2x{)}^3{+}^4{C}_4\right(2x{)}^4]+90x^4{[}^3{C}_0{+}^3{C}_0{+}^3{C}_1(2x\left){+}^3{C}_2\right(2x{)}^2{+}^3{C}_3\left(2x{)}^3\right]-270x^6{[}^2{C}_0{+}^2{C}_1\left(2x\right){+}^2{C}_2(2x{)}^2+5\times 81x^8(1+2x)-243x^{10}]$

$=10+10x+10\times 4x^2+10x8x^3+5\times 16x^4+32x^5-15x^2-120x^3$

$-180x^4+480x^5-240x^6+90x^4+540x^5+1080x^6+720x^7-270x^7-270x^6-1080x^7-1080x^7-1080x^8+405x^8+810x^9-243x^{10}$

$=1+10x+25x^2-40x^3-190x^4+92x^5+570x^6-360x^7-675x^8+810x^9-243x^{10}$

$\left(ix\right){(x+1-\frac{1}{x})}^3$

Let y =x+1, then

${(x+1-\frac{1}{x})}^3={(y-\frac{1}{x})}^3$

The expansion of $(x+y{)}^n$ has n +1 terms so the expansion of ${(y-\frac{1}{x})}^3$ has 4 terms.

Using binomial theorem to expand, we get

${(y-\frac{1}{x})}^3{=}^3{C}_0{y}^3{\left(\frac{1}{x}\right)}^0{-}^3{C}_1{y}^2\left(\frac{1}{x}\right){+}^3{C}_{2}y{\left(\frac{1}{x}\right)}^2{-}^3{C}_3{y}^0{\left(\frac{1}{x}\right)}^3$

$=y^3-3y^2\times \frac{1}{x}+3y\times \frac{1}{x^2}-\frac{1}{x^3}$

Putting y =x+1, we get

${(x+1-\frac{1}{x})}^3=(x+1{)}^3-3(x+1{)}^2\times \frac{1}{x}+3(x+1)\times \frac{1}{x^2}-\frac{1}{x^3}$

$=x^3+1+3x^2+3x-3x-\frac{3}{x}-6+\frac{3}{x}+\frac{3}{x^2}-\frac{1}{x^3}$

$=x^3+3x^2-5+\frac{3}{x^2}-\frac{1}{x^3}$

$\left(x\right)(1-2x+3x^2{)}^3$

Let y =1-2x, then

$(1-2x+3x^2{)}^3=(y+3x^2{)}^3$

The expansion of $(x+y{)}^n$ has n+1 terms so the expansion of $(y+3x^2{)}^3$ has 4 terms.

Using binomial theorem to expand, we get:

$(y+3x^2{)}^3{=}^3{C}_0{y}^3(3x^2{)}^0{+}^3{C}_1{y}^2\left(3x^2{)}^1{+}^3{C}_{2}y\right(3x^2{)}^2{+}^3{C}_3{y}^0(3x^2{)}^3$

$=y^3+3y^2\left(3x^2\right)+2y\left(9x^2\right)+\left(27x^6\right)$

Substituting =1-2x, we get.

$(1-2x+3x^2{)}^3=(1-2x{)}^3+3(1+4x^2-4x)\left(3x^2\right)+3(1-2x)\left(9x^2\right)+\left(27x^6\right)$

$=1-8x^3-6x+12x^2+9x^2+36x^3+27x^2-54x^3+27x^6$

$=1-6x+21x^2-44x^3+63x^4-54x^5+27x^6$