Expand each of the expressions
1. $(1-2x)^{5}$
[$\displaystyle \small (1-x)^{n}=^{n}C_{0}-^{n}C_{1}x+^{n}C_{2}x^{2}-...$ $\displaystyle \small +(-1)^{n}\times ^{n}C_{n}x^{n}$]
= $\displaystyle \small ^{5}C_{0}-^{5}C_{1}(2x)+^{5}C_{2}(2x)^{2}-^{5}C_{3}(2x)^{3}$$\displaystyle \small +^{5}C_{4}(2x)^{4}-^{5}C_{5}(2x)^{5}$
= $\displaystyle \small 1-(5)(2x)+(10)(2x)^{2}-(10)(2x)^{3}$ $\displaystyle \small +(5)(2x)^{4}-(2x)^{5}$
= $\displaystyle \small 1-10x+(10)4x^{2}-(10)8x^{3}$ $\displaystyle \small +(5)16x^{4}-32x^{5}$
= $\displaystyle \small 1-10x+40x^{2}-80x^{3}+80x^{4}-32x^{5}$
2. $\displaystyle \left (\frac{2}{x}-\frac{x}{2} \right)^{5}$
= $\displaystyle \small ^{5}C_{0}\left (\frac{2}{x} \right)^{5}-^{5}C_{1}\left (\frac{2}{x} \right)^{4}\left (\frac{x}{2} \right)$ $\displaystyle \small +^{5}C_{2}\left (\frac{2}{x} \right)^{3}\left (\frac{x}{2} \right)^{2}-^{5}C_{3}\left (\frac{2}{x} \right)^{2}\left (\frac{x}{2} \right)^{3}$ $\displaystyle \small +^{5}C_{4}\left (\frac{2}{x} \right)\left (\frac{x}{2} \right)^{4}-^{5}C_{5}\left (\frac{x}{2} \right)^{5}$
= $\displaystyle \small \left (\frac{2}{x} \right)^{5}-(5)\left (\frac{2}{x} \right)^{4}\left (\frac{x}{2} \right)+(10)\left (\frac{2}{x} \right)^{3}$ $\displaystyle \small \left (\frac{x}{2} \right)^{2}-(10)\left (\frac{2}{x} \right)^{2}\left (\frac{x}{2} \right)^{3}$$\displaystyle \small +(5)\left (\frac{2}{x} \right)\left (\frac{x}{2} \right)^{4}-\left (\frac{x}{2} \right)^{5}$
= $\displaystyle \small \frac{32}{x^{5}}-(5)\left (\frac{16}{x^{4}} \right)\left (\frac{x}{2} \right)+(10)\left (\frac{8}{x^{3}} \right)\left (\frac{x^{2}}{4} \right)$ $\displaystyle \small -(10)\left (\frac{4}{x^{2}} \right)\left (\frac{x^{3}}{8} \right)$$\displaystyle \small +(5)\left (\frac{2}{x} \right)\left (\frac{x^{4}}{16} \right)-\frac{x^{5}}{32}$
= $\displaystyle \small \frac{32}{x^{5}}-(5)\left (\frac{8}{x^{3}} \right)+(10)\left (\frac{2}{x} \right)-(10)\left (\frac{x}{2} \right)$$\displaystyle \small +(5)\left (\frac{x^{3}}{8} \right)-\frac{x^{5}}{32}$
= $\displaystyle \small \frac{32}{x^{5}}-\frac{40}{x^{3}}+\frac{20}{x}-5x$$\displaystyle \small +\frac{5x^{3}}{8}-\frac{x^{5}}{32}$
3. $(2x-3)^{6}$
= $\displaystyle \small ^{6}C_{0}(2x)^{6}-^{6}C_{1}(2x)^{5}(3)+^{6}C_{2}(2x)^{4}(3)^{2}$ $\displaystyle \small -^{6}C_{3}(2x)^{3}(3)^{3}$ $\displaystyle \small +^{6}C_{4}(2x)^{2}(3)^{4}$ $\displaystyle \small -^{6}C_{5}(2x)(3)^{5}+^{6}C_{6}(3)^{6}$
= $\displaystyle \small (2x)^{6}-(6)(2x)^{5}(3)+(15)(2x)^{4}(3)^{2}$ $\displaystyle \small -(20)(2x)^{3}(3)^{3}$ $\displaystyle \small +(15)(2x)^{2}(3)^{4}$ $\displaystyle \small -(6)(2x)(3)^{5}+(3)^{6}$
= $\displaystyle \small 64x^{6}-(6)(32x^{5})(3)+(15)(16x^{4})(9)$ $\displaystyle \small -(20)(8x^{3})(27)$ $\displaystyle \small +(15)(4x^{2})(81)$ $\displaystyle \small -(6)(2x)(243)+729$
= $\displaystyle \small 64x^{6}-576x^{5}+2160x^{4}-4320x^{3}$ $\displaystyle \small +4860x^{2}-2916x+729$
4. $(\frac{x}{3}+\frac{1}{x})^{5}$
= $\displaystyle \small ^{5}C_{0}\left (\frac{x}{3} \right)^{5}+^{5}C_{1}\left (\frac{x}{3} \right)^{4}\left (\frac{1}{x} \right)+^{5}C_{2}\left (\frac{x}{3} \right)^{3}$ $\displaystyle \small \left (\frac{1}{x} \right)^{2}+^{5}C_{3}\left (\frac{x}{3} \right)^{2}\left (\frac{1}{x} \right)^{3}$ $\displaystyle \small +^{5}C_{4}\left (\frac{x}{3} \right)\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +^{5}C_{5}\left (\frac{1}{x} \right)^{5}$
= $\displaystyle \small \left (\frac{x}{3} \right)^{5}+(5)\left (\frac{x}{3} \right)^{4}\left (\frac{1}{x} \right)+(10)\left (\frac{x}{3} \right)^{3}\left (\frac{1}{x} \right)^{2}$ $\displaystyle \small +(10)\left (\frac{x}{3} \right)^{2}\left (\frac{1}{x} \right)^{3}$ $\displaystyle \small +(5)\left (\frac{x}{3} \right)\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +\left (\frac{1}{x} \right)^{5}$
= $\displaystyle \small \frac{x^{5}}{243}+5\left (\frac{x^{4}}{81} \right)\left (\frac{1}{x} \right)+10\left (\frac{x^{3}}{27} \right)\left (\frac{1}{x^{2}} \right)$ $\displaystyle \small +10\left (\frac{x^{2}}{9} \right)\left (\frac{1}{x^{3}} \right)$ $\displaystyle \small +5\left (\frac{x}{3} \right)\left (\frac{1}{x^{4}} \right)$ $\displaystyle \small +\left (\frac{1}{x^{5}} \right)$
= $\displaystyle \small \frac{x^{5}}{243}+\frac{5x^{3}}{81}+\frac{10x}{27}+\frac{10}{9x}$ $\displaystyle \small +\frac{5}{3x^{3}}+\frac{1}{x^{5}}$
5. $\displaystyle \small (x+\frac{1}{x})^{6}$
= $\displaystyle \small ^{6}C_{0}(x)^{6}+^{6}C_{1}(x)^{5}\left (\frac{1}{x} \right)+^{6}C_{2}(x)^{4}\left (\frac{1}{x} \right)^{2}$ $\displaystyle \small +^{6}C_{3}(x)^{3}\left (\frac{1}{x} \right)^{3}+^{6}C_{4}(x)^{2}\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +^{6}C_{5}(x)\left (\frac{1}{x} \right)^{5}+^{6}C_{6}\left (\frac{1}{x} \right)^{6}$
= $\displaystyle \small x^{6}+(6)(x)^{5}\left (\frac{1}{x} \right)+(15)(x)^{4}\left (\frac{1}{x} \right)^{2}$ $\displaystyle \small +(20)(x)^{3}\left (\frac{1}{x} \right)^{3}+(15)(x)^{2}\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +(6)(x)\left (\frac{1}{x} \right)^{5}+\left (\frac{1}{x} \right)^{6}$
= $\displaystyle \small x^{6}+6(x)^{5}\left (\frac{1}{x} \right)+15(x)^{4}\left (\frac{1}{x^{2}} \right)$ $\displaystyle \small +20(x)^{3}\left (\frac{1}{x^{3}} \right)+15(x)^{2}\left (\frac{1}{x^{4}} \right)$ $\displaystyle \small +6(x)\left (\frac{1}{x^{5}} \right)+\frac{1}{x^{6}}$
= $\displaystyle \small x^{6}+6x^{4}+15x^{2}+20+\frac{15}{x^{2}}$ $\displaystyle \small +\frac{6}{x^{4}}+\frac{1}{x^{6}}$
Using binomial theorem, evaluate the following
6. $(96)^{3}$
$\displaystyle \small (100-4)^{3}$
= $\displaystyle \small ^{3}C_{0}(100)^{3}-^{3}C_{1}(100)^{2}(4)+^{3}C_{2}(100)(4)^{2}$ $\displaystyle \small -^{3}C_{3}(4)^{3}$
= $\displaystyle \small (100)^{3}-3(100)^{2}(4)+3(100)(4)^{2}-(4)^{3}$
= $\displaystyle \small 1000000-3(10000)(4)+3(100)(16)-64$
= $\displaystyle \small 1000000-120000+4800-64$
= $\displaystyle \small 884736$
7. $(102)^{5}$
$\displaystyle \small (100+2)^{5}$
= $\displaystyle \small ^{5}C_{0}(100)^{5}+^{5}C_{1}(100)^{4}(2)+^{5}C_{2}(100)^{3}(2)^{2}$ $\displaystyle \small +^{5}C_{3}(100)^{2}(2)^{3}$ $\displaystyle \small +^{5}C_{4}(100)(2)^{4}+^{5}C_{5}(2)^{5}$
= $\displaystyle \small (100)^{5}+5(100)^{4}(2)+10(100)^{3}(2)^{2}$ $\displaystyle \small +10(100)^{2}(2)^{3}$ $\displaystyle \small +5(100)(2)^{4}+(2)^{5}$
= $\displaystyle \small 10000000000+5(100000000)(2)$ $\displaystyle \small +10(1000000)(4)+10(10000)(8)$ $\displaystyle \small +5(100)(16)+32$
= $\displaystyle \small 10000000000+1000000000+40000000$ $\displaystyle \small +800000$ $\displaystyle \small +8000+32$
= $\displaystyle \small 11040808032$
8. $(101)^{4}$
$\displaystyle \small (100+1)^{4}$
= $\displaystyle \small ^{4}C_{0}(100)^{4}+^{4}C_{1}(100)^{3}+^{4}C_{2}(100)^{2}$ $\displaystyle \small +^{4}C_{3}(100)+^{4}C_{4}$
= $\displaystyle \small (100)^{4}+4(100)^{3}+6(100)^{2}$ $\displaystyle \small +4(100)+1$
= $\displaystyle \small 100000000+4(1000000)+6(10000)$ $\displaystyle \small +4(100)$ $\displaystyle \small +1$
= $\displaystyle \small 100000000+4000000+60000+400+1$
= $\displaystyle \small 104060401$
9. $(99)^{5}$
$\displaystyle \small (100-1)^{5}$
= $\displaystyle \small ^{5}C_{0}(100)^{5}-^{5}C_{1}(100)^{4}+^{5}C_{2}(100)^{3}$ $\displaystyle \small -^{5}C_{3}(100)^{2}+^{5}C_{4}(100)-^{5}C_{5}$
= $\displaystyle \small (100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}$ $\displaystyle \small +5(100)-1$
= $\displaystyle \small 10000000000-5(100000000)$ $\displaystyle \small +10(1000000)$ $\displaystyle \small -10(10000)+5(100)$ $\displaystyle \small -1$
= $\displaystyle \small 10000000000-500000000+10000000$ $\displaystyle \small -100000+500-1$
= $\displaystyle \small 9509900499$
10. Using Binomial theorem, indicate which number is larger, $(1.1)^{10000}$ or $1000$.
$\displaystyle \small (1+0.1)^{10000}$
= $\displaystyle \small ^{10000}C_{0}(1)^{10000}+^{10000}C_{1}(1)^{999}(0.1)$ $\displaystyle \small +^{10000}C_{2}(1)^{998}(0.1)^{2}+...$
= $\displaystyle \small (1)^{10000}+10000(1)^{999}(0.1)+$ other positive numbers
= $\displaystyle \small 1+10000(0.1)+$ other positive numbers
= $\displaystyle \small 1+1000+$ other positive numbers
which is greater than 1000
∴ $\displaystyle \small (1.1)^{10000}>1000$
11. Find $(a+b)^{4}-(a-b)^{4}$. Hence evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$
$\displaystyle \small (a+b)^{4}-(a-b)^{4}$
= [$\displaystyle \small ^{4}C_{0}a^{4}+^{4}C_{1}a^{3}b+^{4}C_{2}a^{2}b^{2}$ $\displaystyle \small +^{4}C_{3}ab^{3}$ $\displaystyle \small +^{4}C_{4}b^{4}$]-[$\displaystyle \small ^{4}C_{0}a^{4}-^{4}C_{1}a^{3}b$ $\displaystyle \small +^{4}C_{2}a^{2}b^{2}$ $\displaystyle \small -^{4}C_{3}ab^{3}+^{4}C_{4}b^{4}$]
= [$\displaystyle \small a^{4}+4a^{3}b$ $\displaystyle \small +6a^{2}b^{2}$ $\displaystyle \small +4ab^{3}+b^{4}$]-[$\displaystyle \small a^{4}-^4a^{3}b+6a^{2}b^{2}$
$\displaystyle \small -4ab^{3}+b^{4}$]
= $\displaystyle \small a^{4}+4a^{3}b+6a^{2}b^{2}$ $\displaystyle \small +4ab^{3}+b^{4}$$\displaystyle \small -a^{4}+^4a^{3}b$ $\displaystyle \small -6a^{2}b^{2}$ $\displaystyle \small +4ab^{3}-b^{4}$
= $\displaystyle \small 8a^{3}b+8ab^{3}$
= $\displaystyle \small 8ab(a^{2}+b^{2})$
Substitute $\displaystyle \small a=\sqrt{3}$ and $\displaystyle \small b=\sqrt{2}$
$\displaystyle \small (\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$
= $\displaystyle \small 8\sqrt{3}\sqrt{2}((\sqrt{3})^{2}+(\sqrt{2})^{2})$
= $\displaystyle \small 8\sqrt{6}(3+2)$
= $\displaystyle \small 8\sqrt{6}(5)$
= $\displaystyle \small 40\sqrt{6}$
12. Find $(x+1)^{6}+(x-1)^{6}$. Hence or otherwise evaluate $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$
$\displaystyle \small (x+1)^{6}+(x-1)^{6}$
= [$\displaystyle \small ^{6}C_{0}x^{6}+^{6}C_{1}x^{5}+^{6}C_{2}x^{4} +^{6}C_{3}x^{3}+^{6}C_{4}x^{2}$ $\displaystyle \small +^{6}C_{5}x+^{6}C_{6}]$ $\displaystyle \small +[^{6}C_{0}x^{6}-^{6}C_{1}x^{5}+^{6}C_{2}x^{4}$ $\displaystyle \small -^{6}C_{3}x^{3}+^{6}C_{4}x^{2}-^{6}C_{5}x+^{6}C_{6}$]
= [$\displaystyle \small x^{6}+6x^{5}+15x^{4}$ $\displaystyle \small +20x^{3}+15x^{2}+6x+1$]+[$\displaystyle \small x^{6}-6x^{5}+15x^{4}$ $\displaystyle \small -20x^{3}+15x^{2}-6x+1$]
= $\displaystyle \small 2x^{6}+30x^{4}+30x^{2}+2$
= $\displaystyle \small 2(x^{6}+15x^{4}+15x^{2}+1)$
Substitute $\displaystyle \small x=\sqrt{2}$
= $\displaystyle \small 2((\sqrt{2})^{6}+15(\sqrt{2})^{4}+15(\sqrt{2})^{2}+1)$
= $\displaystyle \small 2(8+15(4)+15(2)+1)$
= $\displaystyle \small 2(8+60+30+1)=2*99=198$
13. Show that $9^{n+1}-8n-9$ is divisible by $64$, whenever n is a positive integer.
Let $\displaystyle \small 9^{n+1}=(1+8)^{n+1}$
= $\displaystyle \small ^{n+1}C_{0}+^{n+1}C_{1}(8)+^{n+1}C_{2}(8)^{2}$ $\displaystyle \small +^{n+1}C_{3}(8)^{3}$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n+1}$
= $\displaystyle \small 1+(n+1)(8)+[^{n+1}C_{2}(8)^{2}$ $\displaystyle \small +^{n+1}C_{3}(8)^{3}$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n+1}]$
= $\displaystyle \small 1+8n+8+[^{n+1}C_{2}(8)^{2}$ $\displaystyle \small +^{n+1}C_{3}(8)^{3}$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n+1}]$
= $\displaystyle \small 8n+9+(8)^{2}[^{n+1}C_{2}$ $\displaystyle \small +^{n+1}C_{3}(8)+...+^{n+1}C_{n+1}(8)^{n-1}]$
$\displaystyle \small 9^{n+1}=8n+9+64[^{n+1}C_{2}$ $\displaystyle \small +^{n+1}C_{3}(8)$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n-1}]$
$\displaystyle \small 9^{n+1}-8n-9=64[^{n+1}C_{2}$ $\displaystyle \small +^{n+1}C_{3}(8)$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n-1}]$
which shows that $\displaystyle \small 9^{n+1}-8n-9$ is divisible by $\displaystyle \small 64$
14. Prove that $\displaystyle \sum_{r=0}^{n}3^{r}(^{n}C_{r})=4^{n}$
$\displaystyle \small \sum_{r=0}^{n}3^{r}(^{n}C_{r})$
= $\displaystyle \small 3^{0}(^{n}C_{0})+3^{1}(^{n}C_{1})+3^{2}(^{n}C_{2})+...$ $\displaystyle \small +3^{n}(^{n}C_{n})$
= $\displaystyle \small ^{n}C_{0}(1)^{n}3^{0}+^{n}C_{1}(1)^{n-1}3^{1}+^{n}C_{2}(1)^{n-2}3^{2}$ $\displaystyle \small +...$ $\displaystyle \small +^{n}C_{n}(1)^{0}3^{n}$
= $\displaystyle \small (1+3)^{n}$
= $\displaystyle \small 4^{n}$
1. $(1-2x)^{5}$
[$\displaystyle \small (1-x)^{n}=^{n}C_{0}-^{n}C_{1}x+^{n}C_{2}x^{2}-...$ $\displaystyle \small +(-1)^{n}\times ^{n}C_{n}x^{n}$]
= $\displaystyle \small ^{5}C_{0}-^{5}C_{1}(2x)+^{5}C_{2}(2x)^{2}-^{5}C_{3}(2x)^{3}$$\displaystyle \small +^{5}C_{4}(2x)^{4}-^{5}C_{5}(2x)^{5}$
= $\displaystyle \small 1-(5)(2x)+(10)(2x)^{2}-(10)(2x)^{3}$ $\displaystyle \small +(5)(2x)^{4}-(2x)^{5}$
= $\displaystyle \small 1-10x+(10)4x^{2}-(10)8x^{3}$ $\displaystyle \small +(5)16x^{4}-32x^{5}$
= $\displaystyle \small 1-10x+40x^{2}-80x^{3}+80x^{4}-32x^{5}$
2. $\displaystyle \left (\frac{2}{x}-\frac{x}{2} \right)^{5}$
= $\displaystyle \small ^{5}C_{0}\left (\frac{2}{x} \right)^{5}-^{5}C_{1}\left (\frac{2}{x} \right)^{4}\left (\frac{x}{2} \right)$ $\displaystyle \small +^{5}C_{2}\left (\frac{2}{x} \right)^{3}\left (\frac{x}{2} \right)^{2}-^{5}C_{3}\left (\frac{2}{x} \right)^{2}\left (\frac{x}{2} \right)^{3}$ $\displaystyle \small +^{5}C_{4}\left (\frac{2}{x} \right)\left (\frac{x}{2} \right)^{4}-^{5}C_{5}\left (\frac{x}{2} \right)^{5}$
= $\displaystyle \small \left (\frac{2}{x} \right)^{5}-(5)\left (\frac{2}{x} \right)^{4}\left (\frac{x}{2} \right)+(10)\left (\frac{2}{x} \right)^{3}$ $\displaystyle \small \left (\frac{x}{2} \right)^{2}-(10)\left (\frac{2}{x} \right)^{2}\left (\frac{x}{2} \right)^{3}$$\displaystyle \small +(5)\left (\frac{2}{x} \right)\left (\frac{x}{2} \right)^{4}-\left (\frac{x}{2} \right)^{5}$
= $\displaystyle \small \frac{32}{x^{5}}-(5)\left (\frac{16}{x^{4}} \right)\left (\frac{x}{2} \right)+(10)\left (\frac{8}{x^{3}} \right)\left (\frac{x^{2}}{4} \right)$ $\displaystyle \small -(10)\left (\frac{4}{x^{2}} \right)\left (\frac{x^{3}}{8} \right)$$\displaystyle \small +(5)\left (\frac{2}{x} \right)\left (\frac{x^{4}}{16} \right)-\frac{x^{5}}{32}$
= $\displaystyle \small \frac{32}{x^{5}}-(5)\left (\frac{8}{x^{3}} \right)+(10)\left (\frac{2}{x} \right)-(10)\left (\frac{x}{2} \right)$$\displaystyle \small +(5)\left (\frac{x^{3}}{8} \right)-\frac{x^{5}}{32}$
= $\displaystyle \small \frac{32}{x^{5}}-\frac{40}{x^{3}}+\frac{20}{x}-5x$$\displaystyle \small +\frac{5x^{3}}{8}-\frac{x^{5}}{32}$
3. $(2x-3)^{6}$
= $\displaystyle \small ^{6}C_{0}(2x)^{6}-^{6}C_{1}(2x)^{5}(3)+^{6}C_{2}(2x)^{4}(3)^{2}$ $\displaystyle \small -^{6}C_{3}(2x)^{3}(3)^{3}$ $\displaystyle \small +^{6}C_{4}(2x)^{2}(3)^{4}$ $\displaystyle \small -^{6}C_{5}(2x)(3)^{5}+^{6}C_{6}(3)^{6}$
= $\displaystyle \small (2x)^{6}-(6)(2x)^{5}(3)+(15)(2x)^{4}(3)^{2}$ $\displaystyle \small -(20)(2x)^{3}(3)^{3}$ $\displaystyle \small +(15)(2x)^{2}(3)^{4}$ $\displaystyle \small -(6)(2x)(3)^{5}+(3)^{6}$
= $\displaystyle \small 64x^{6}-(6)(32x^{5})(3)+(15)(16x^{4})(9)$ $\displaystyle \small -(20)(8x^{3})(27)$ $\displaystyle \small +(15)(4x^{2})(81)$ $\displaystyle \small -(6)(2x)(243)+729$
= $\displaystyle \small 64x^{6}-576x^{5}+2160x^{4}-4320x^{3}$ $\displaystyle \small +4860x^{2}-2916x+729$
4. $(\frac{x}{3}+\frac{1}{x})^{5}$
= $\displaystyle \small ^{5}C_{0}\left (\frac{x}{3} \right)^{5}+^{5}C_{1}\left (\frac{x}{3} \right)^{4}\left (\frac{1}{x} \right)+^{5}C_{2}\left (\frac{x}{3} \right)^{3}$ $\displaystyle \small \left (\frac{1}{x} \right)^{2}+^{5}C_{3}\left (\frac{x}{3} \right)^{2}\left (\frac{1}{x} \right)^{3}$ $\displaystyle \small +^{5}C_{4}\left (\frac{x}{3} \right)\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +^{5}C_{5}\left (\frac{1}{x} \right)^{5}$
= $\displaystyle \small \left (\frac{x}{3} \right)^{5}+(5)\left (\frac{x}{3} \right)^{4}\left (\frac{1}{x} \right)+(10)\left (\frac{x}{3} \right)^{3}\left (\frac{1}{x} \right)^{2}$ $\displaystyle \small +(10)\left (\frac{x}{3} \right)^{2}\left (\frac{1}{x} \right)^{3}$ $\displaystyle \small +(5)\left (\frac{x}{3} \right)\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +\left (\frac{1}{x} \right)^{5}$
= $\displaystyle \small \frac{x^{5}}{243}+5\left (\frac{x^{4}}{81} \right)\left (\frac{1}{x} \right)+10\left (\frac{x^{3}}{27} \right)\left (\frac{1}{x^{2}} \right)$ $\displaystyle \small +10\left (\frac{x^{2}}{9} \right)\left (\frac{1}{x^{3}} \right)$ $\displaystyle \small +5\left (\frac{x}{3} \right)\left (\frac{1}{x^{4}} \right)$ $\displaystyle \small +\left (\frac{1}{x^{5}} \right)$
= $\displaystyle \small \frac{x^{5}}{243}+\frac{5x^{3}}{81}+\frac{10x}{27}+\frac{10}{9x}$ $\displaystyle \small +\frac{5}{3x^{3}}+\frac{1}{x^{5}}$
5. $\displaystyle \small (x+\frac{1}{x})^{6}$
= $\displaystyle \small ^{6}C_{0}(x)^{6}+^{6}C_{1}(x)^{5}\left (\frac{1}{x} \right)+^{6}C_{2}(x)^{4}\left (\frac{1}{x} \right)^{2}$ $\displaystyle \small +^{6}C_{3}(x)^{3}\left (\frac{1}{x} \right)^{3}+^{6}C_{4}(x)^{2}\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +^{6}C_{5}(x)\left (\frac{1}{x} \right)^{5}+^{6}C_{6}\left (\frac{1}{x} \right)^{6}$
= $\displaystyle \small x^{6}+(6)(x)^{5}\left (\frac{1}{x} \right)+(15)(x)^{4}\left (\frac{1}{x} \right)^{2}$ $\displaystyle \small +(20)(x)^{3}\left (\frac{1}{x} \right)^{3}+(15)(x)^{2}\left (\frac{1}{x} \right)^{4}$ $\displaystyle \small +(6)(x)\left (\frac{1}{x} \right)^{5}+\left (\frac{1}{x} \right)^{6}$
= $\displaystyle \small x^{6}+6(x)^{5}\left (\frac{1}{x} \right)+15(x)^{4}\left (\frac{1}{x^{2}} \right)$ $\displaystyle \small +20(x)^{3}\left (\frac{1}{x^{3}} \right)+15(x)^{2}\left (\frac{1}{x^{4}} \right)$ $\displaystyle \small +6(x)\left (\frac{1}{x^{5}} \right)+\frac{1}{x^{6}}$
= $\displaystyle \small x^{6}+6x^{4}+15x^{2}+20+\frac{15}{x^{2}}$ $\displaystyle \small +\frac{6}{x^{4}}+\frac{1}{x^{6}}$
Using binomial theorem, evaluate the following
6. $(96)^{3}$
$\displaystyle \small (100-4)^{3}$
= $\displaystyle \small ^{3}C_{0}(100)^{3}-^{3}C_{1}(100)^{2}(4)+^{3}C_{2}(100)(4)^{2}$ $\displaystyle \small -^{3}C_{3}(4)^{3}$
= $\displaystyle \small (100)^{3}-3(100)^{2}(4)+3(100)(4)^{2}-(4)^{3}$
= $\displaystyle \small 1000000-3(10000)(4)+3(100)(16)-64$
= $\displaystyle \small 1000000-120000+4800-64$
= $\displaystyle \small 884736$
7. $(102)^{5}$
$\displaystyle \small (100+2)^{5}$
= $\displaystyle \small ^{5}C_{0}(100)^{5}+^{5}C_{1}(100)^{4}(2)+^{5}C_{2}(100)^{3}(2)^{2}$ $\displaystyle \small +^{5}C_{3}(100)^{2}(2)^{3}$ $\displaystyle \small +^{5}C_{4}(100)(2)^{4}+^{5}C_{5}(2)^{5}$
= $\displaystyle \small (100)^{5}+5(100)^{4}(2)+10(100)^{3}(2)^{2}$ $\displaystyle \small +10(100)^{2}(2)^{3}$ $\displaystyle \small +5(100)(2)^{4}+(2)^{5}$
= $\displaystyle \small 10000000000+5(100000000)(2)$ $\displaystyle \small +10(1000000)(4)+10(10000)(8)$ $\displaystyle \small +5(100)(16)+32$
= $\displaystyle \small 10000000000+1000000000+40000000$ $\displaystyle \small +800000$ $\displaystyle \small +8000+32$
= $\displaystyle \small 11040808032$
8. $(101)^{4}$
$\displaystyle \small (100+1)^{4}$
= $\displaystyle \small ^{4}C_{0}(100)^{4}+^{4}C_{1}(100)^{3}+^{4}C_{2}(100)^{2}$ $\displaystyle \small +^{4}C_{3}(100)+^{4}C_{4}$
= $\displaystyle \small (100)^{4}+4(100)^{3}+6(100)^{2}$ $\displaystyle \small +4(100)+1$
= $\displaystyle \small 100000000+4(1000000)+6(10000)$ $\displaystyle \small +4(100)$ $\displaystyle \small +1$
= $\displaystyle \small 100000000+4000000+60000+400+1$
= $\displaystyle \small 104060401$
9. $(99)^{5}$
$\displaystyle \small (100-1)^{5}$
= $\displaystyle \small ^{5}C_{0}(100)^{5}-^{5}C_{1}(100)^{4}+^{5}C_{2}(100)^{3}$ $\displaystyle \small -^{5}C_{3}(100)^{2}+^{5}C_{4}(100)-^{5}C_{5}$
= $\displaystyle \small (100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}$ $\displaystyle \small +5(100)-1$
= $\displaystyle \small 10000000000-5(100000000)$ $\displaystyle \small +10(1000000)$ $\displaystyle \small -10(10000)+5(100)$ $\displaystyle \small -1$
= $\displaystyle \small 10000000000-500000000+10000000$ $\displaystyle \small -100000+500-1$
= $\displaystyle \small 9509900499$
10. Using Binomial theorem, indicate which number is larger, $(1.1)^{10000}$ or $1000$.
$\displaystyle \small (1+0.1)^{10000}$
= $\displaystyle \small ^{10000}C_{0}(1)^{10000}+^{10000}C_{1}(1)^{999}(0.1)$ $\displaystyle \small +^{10000}C_{2}(1)^{998}(0.1)^{2}+...$
= $\displaystyle \small (1)^{10000}+10000(1)^{999}(0.1)+$ other positive numbers
= $\displaystyle \small 1+10000(0.1)+$ other positive numbers
= $\displaystyle \small 1+1000+$ other positive numbers
which is greater than 1000
∴ $\displaystyle \small (1.1)^{10000}>1000$
11. Find $(a+b)^{4}-(a-b)^{4}$. Hence evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$
$\displaystyle \small (a+b)^{4}-(a-b)^{4}$
= [$\displaystyle \small ^{4}C_{0}a^{4}+^{4}C_{1}a^{3}b+^{4}C_{2}a^{2}b^{2}$ $\displaystyle \small +^{4}C_{3}ab^{3}$ $\displaystyle \small +^{4}C_{4}b^{4}$]-[$\displaystyle \small ^{4}C_{0}a^{4}-^{4}C_{1}a^{3}b$ $\displaystyle \small +^{4}C_{2}a^{2}b^{2}$ $\displaystyle \small -^{4}C_{3}ab^{3}+^{4}C_{4}b^{4}$]
= [$\displaystyle \small a^{4}+4a^{3}b$ $\displaystyle \small +6a^{2}b^{2}$ $\displaystyle \small +4ab^{3}+b^{4}$]-[$\displaystyle \small a^{4}-^4a^{3}b+6a^{2}b^{2}$
$\displaystyle \small -4ab^{3}+b^{4}$]
= $\displaystyle \small a^{4}+4a^{3}b+6a^{2}b^{2}$ $\displaystyle \small +4ab^{3}+b^{4}$$\displaystyle \small -a^{4}+^4a^{3}b$ $\displaystyle \small -6a^{2}b^{2}$ $\displaystyle \small +4ab^{3}-b^{4}$
= $\displaystyle \small 8a^{3}b+8ab^{3}$
= $\displaystyle \small 8ab(a^{2}+b^{2})$
Substitute $\displaystyle \small a=\sqrt{3}$ and $\displaystyle \small b=\sqrt{2}$
$\displaystyle \small (\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$
= $\displaystyle \small 8\sqrt{3}\sqrt{2}((\sqrt{3})^{2}+(\sqrt{2})^{2})$
= $\displaystyle \small 8\sqrt{6}(3+2)$
= $\displaystyle \small 8\sqrt{6}(5)$
= $\displaystyle \small 40\sqrt{6}$
12. Find $(x+1)^{6}+(x-1)^{6}$. Hence or otherwise evaluate $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$
$\displaystyle \small (x+1)^{6}+(x-1)^{6}$
= [$\displaystyle \small ^{6}C_{0}x^{6}+^{6}C_{1}x^{5}+^{6}C_{2}x^{4} +^{6}C_{3}x^{3}+^{6}C_{4}x^{2}$ $\displaystyle \small +^{6}C_{5}x+^{6}C_{6}]$ $\displaystyle \small +[^{6}C_{0}x^{6}-^{6}C_{1}x^{5}+^{6}C_{2}x^{4}$ $\displaystyle \small -^{6}C_{3}x^{3}+^{6}C_{4}x^{2}-^{6}C_{5}x+^{6}C_{6}$]
= [$\displaystyle \small x^{6}+6x^{5}+15x^{4}$ $\displaystyle \small +20x^{3}+15x^{2}+6x+1$]+[$\displaystyle \small x^{6}-6x^{5}+15x^{4}$ $\displaystyle \small -20x^{3}+15x^{2}-6x+1$]
= $\displaystyle \small 2x^{6}+30x^{4}+30x^{2}+2$
= $\displaystyle \small 2(x^{6}+15x^{4}+15x^{2}+1)$
Substitute $\displaystyle \small x=\sqrt{2}$
= $\displaystyle \small 2((\sqrt{2})^{6}+15(\sqrt{2})^{4}+15(\sqrt{2})^{2}+1)$
= $\displaystyle \small 2(8+15(4)+15(2)+1)$
= $\displaystyle \small 2(8+60+30+1)=2*99=198$
13. Show that $9^{n+1}-8n-9$ is divisible by $64$, whenever n is a positive integer.
Let $\displaystyle \small 9^{n+1}=(1+8)^{n+1}$
= $\displaystyle \small ^{n+1}C_{0}+^{n+1}C_{1}(8)+^{n+1}C_{2}(8)^{2}$ $\displaystyle \small +^{n+1}C_{3}(8)^{3}$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n+1}$
= $\displaystyle \small 1+(n+1)(8)+[^{n+1}C_{2}(8)^{2}$ $\displaystyle \small +^{n+1}C_{3}(8)^{3}$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n+1}]$
= $\displaystyle \small 1+8n+8+[^{n+1}C_{2}(8)^{2}$ $\displaystyle \small +^{n+1}C_{3}(8)^{3}$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n+1}]$
= $\displaystyle \small 8n+9+(8)^{2}[^{n+1}C_{2}$ $\displaystyle \small +^{n+1}C_{3}(8)+...+^{n+1}C_{n+1}(8)^{n-1}]$
$\displaystyle \small 9^{n+1}=8n+9+64[^{n+1}C_{2}$ $\displaystyle \small +^{n+1}C_{3}(8)$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n-1}]$
$\displaystyle \small 9^{n+1}-8n-9=64[^{n+1}C_{2}$ $\displaystyle \small +^{n+1}C_{3}(8)$ $\displaystyle \small +...+^{n+1}C_{n+1}(8)^{n-1}]$
which shows that $\displaystyle \small 9^{n+1}-8n-9$ is divisible by $\displaystyle \small 64$
14. Prove that $\displaystyle \sum_{r=0}^{n}3^{r}(^{n}C_{r})=4^{n}$
$\displaystyle \small \sum_{r=0}^{n}3^{r}(^{n}C_{r})$
= $\displaystyle \small 3^{0}(^{n}C_{0})+3^{1}(^{n}C_{1})+3^{2}(^{n}C_{2})+...$ $\displaystyle \small +3^{n}(^{n}C_{n})$
= $\displaystyle \small ^{n}C_{0}(1)^{n}3^{0}+^{n}C_{1}(1)^{n-1}3^{1}+^{n}C_{2}(1)^{n-2}3^{2}$ $\displaystyle \small +...$ $\displaystyle \small +^{n}C_{n}(1)^{0}3^{n}$
= $\displaystyle \small (1+3)^{n}$
= $\displaystyle \small 4^{n}$
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