Class 11: Binomial Theorem

$\displaystyle \textbf{Question 1: }\text{Using the binomial theorem, write down the expansions of the following:}$
$\displaystyle \text{i) }(2x+3y)^5\qquad\text{ii) }(2x-3y)^4\qquad\text{iii) }\left(x-\frac{1}{x}\right)^6$
$\displaystyle \text{iv) }(1-3x)^7\qquad\text{v) }\left(ax-\frac{b}{x}\right)^6$
$\displaystyle \text{vi) }\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)^6\qquad\text{vii) }\left(\sqrt[3]{x}-\sqrt[3]{a}\right)^6$
$\displaystyle \text{viii) }(1+2x-3x^2)^5\qquad\text{ix) }\left(x+1-\frac{1}{x}\right)^3$
$\displaystyle \text{x) }(1-2x+3x^2)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }(2x+3y)^5$
$\displaystyle ={}^5C_0(2x)^5+{}^5C_1(2x)^4(3y)+{}^5C_2(2x)^3(3y)^2$
$\displaystyle \qquad+{}^5C_3(2x)^2(3y)^3+{}^5C_4(2x)(3y)^4+{}^5C_5(3y)^5$
$\displaystyle =32x^5+240x^4y+720x^3y^2+1080x^2y^3+810xy^4+243y^5$

$\displaystyle \text{ii) }(2x-3y)^4$
$\displaystyle ={}^4C_0(2x)^4-{}^4C_1(2x)^3(3y)+{}^4C_2(2x)^2(3y)^2$
$\displaystyle \qquad-{}^4C_3(2x)(3y)^3+{}^4C_4(3y)^4$
$\displaystyle =16x^4-96x^3y+216x^2y^2-216xy^3+81y^4$

$\displaystyle \text{iii) }\left(x-\frac{1}{x}\right)^6$
$\displaystyle =x^6-6x^5\left(\frac{1}{x}\right)+15x^4\left(\frac{1}{x}\right)^2$
$\displaystyle \qquad-20x^3\left(\frac{1}{x}\right)^3+15x^2\left(\frac{1}{x}\right)^4$
$\displaystyle \qquad-6x\left(\frac{1}{x}\right)^5+\left(\frac{1}{x}\right)^6$
$\displaystyle =x^6-6x^4+15x^2-20+\frac{15}{x^2}-\frac{6}{x^4}+\frac{1}{x^6},\quad x\ne0$

$\displaystyle \text{iv) }(1-3x)^7$
$\displaystyle =1-{}^7C_1(3x)+{}^7C_2(3x)^2-{}^7C_3(3x)^3+{}^7C_4(3x)^4$
$\displaystyle \qquad-{}^7C_5(3x)^5+{}^7C_6(3x)^6-{}^7C_7(3x)^7$
$\displaystyle =1-21x+189x^2-945x^3+2835x^4-5103x^5+5103x^6-2187x^7$

$\displaystyle \text{v) }\left(ax-\frac{b}{x}\right)^6$
$\displaystyle =(ax)^6-6(ax)^5\left(\frac{b}{x}\right)+15(ax)^4\left(\frac{b}{x}\right)^2$
$\displaystyle \qquad-20(ax)^3\left(\frac{b}{x}\right)^3+15(ax)^2\left(\frac{b}{x}\right)^4$
$\displaystyle \qquad-6(ax)\left(\frac{b}{x}\right)^5+\left(\frac{b}{x}\right)^6$
$\displaystyle =a^6x^6-6a^5bx^4+15a^4b^2x^2-20a^3b^3$
$\displaystyle \qquad+\frac{15a^2b^4}{x^2}-\frac{6ab^5}{x^4}+\frac{b^6}{x^6},\quad x\ne0$

$\displaystyle \text{vi) }\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)^6$
$\displaystyle =\left(\sqrt{\frac{x}{a}}\right)^6-6\left(\sqrt{\frac{x}{a}}\right)^5\sqrt{\frac{a}{x}}$
$\displaystyle \qquad+15\left(\sqrt{\frac{x}{a}}\right)^4\left(\sqrt{\frac{a}{x}}\right)^2$
$\displaystyle \qquad-20\left(\sqrt{\frac{x}{a}}\right)^3\left(\sqrt{\frac{a}{x}}\right)^3$
$\displaystyle \qquad+15\left(\sqrt{\frac{x}{a}}\right)^2\left(\sqrt{\frac{a}{x}}\right)^4$
$\displaystyle \qquad-6\sqrt{\frac{x}{a}}\left(\sqrt{\frac{a}{x}}\right)^5+\left(\sqrt{\frac{a}{x}}\right)^6$
$\displaystyle =\frac{x^3}{a^3}-\frac{6x^2}{a^2}+\frac{15x}{a}-20+\frac{15a}{x}-\frac{6a^2}{x^2}+\frac{a^3}{x^3}$

$\displaystyle \text{vii) }\left(\sqrt[3]{x}-\sqrt[3]{a}\right)^6$
$\displaystyle =x^2-6x^{\frac{5}{3}}a^{\frac{1}{3}}+15x^{\frac{4}{3}}a^{\frac{2}{3}}-20ax$
$\displaystyle \qquad+15x^{\frac{2}{3}}a^{\frac{4}{3}}-6x^{\frac{1}{3}}a^{\frac{5}{3}}+a^2$

$\displaystyle \text{viii) }(1+2x-3x^2)^5$
$\displaystyle =1+10x+25x^2-40x^3-190x^4+92x^5$
$\displaystyle \qquad+570x^6-360x^7-675x^8+810x^9-243x^{10}$

$\displaystyle \text{ix) }\left(x+1-\frac{1}{x}\right)^3$
$\displaystyle =(x+1)^3-\frac{3(x+1)^2}{x}+\frac{3(x+1)}{x^2}-\frac{1}{x^3}$
$\displaystyle =x^3+3x^2+3x+1-3x-6-\frac{3}{x}$
$\displaystyle \qquad+\frac{3}{x}+\frac{3}{x^2}-\frac{1}{x^3}$
$\displaystyle =x^3+3x^2-5+\frac{3}{x^2}-\frac{1}{x^3},\quad x\ne0$

$\displaystyle \text{x) }(1-2x+3x^2)^3$
$\displaystyle =(1-2x)^3+3(1-2x)^2(3x^2)+3(1-2x)(3x^2)^2+(3x^2)^3$
$\displaystyle =(1-6x+12x^2-8x^3)+(9x^2-36x^3+36x^4)$
$\displaystyle \qquad+(27x^4-54x^5)+27x^6$
$\displaystyle =1-6x+21x^2-44x^3+63x^4-54x^5+27x^6$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Evaluate the following:}$
$\displaystyle \text{i) }\left(\sqrt{x+1}+\sqrt{x-1}\right)^6+\left(\sqrt{x+1}-\sqrt{x-1}\right)^6$
$\displaystyle \text{ii) }\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6$
$\displaystyle \text{iii) }(1+2\sqrt{x})^5-(1-2\sqrt{x})^5$
$\displaystyle \text{iv) }(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$
$\displaystyle \text{v) }(3+\sqrt{2})^5-(3-\sqrt{2})^5$
$\displaystyle \text{vi) }(2+\sqrt{3})^7+(2-\sqrt{3})^7$
$\displaystyle \text{vii) }(\sqrt{3}+1)^5-(\sqrt{3}-1)^5$
$\displaystyle \text{viii) }(0.99)^5+(1.01)^5$
$\displaystyle \text{ix) }(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6$
$\displaystyle \text{x) }\left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }\left(\sqrt{x+1}+\sqrt{x-1}\right)^6+\left(\sqrt{x+1}-\sqrt{x-1}\right)^6$
$\displaystyle =2\left[{}^6C_0(x+1)^3+{}^6C_2(x+1)^2(x-1)\right.$
$\displaystyle \left.\qquad+{}^6C_4(x+1)(x-1)^2+{}^6C_6(x-1)^3\right]$
$\displaystyle =2\left[(x+1)^3+15(x+1)^2(x-1)\right.$
$\displaystyle \left.\qquad+15(x+1)(x-1)^2+(x-1)^3\right]$
$\displaystyle =2\left[2x^3+6x+15(x^3+x^2-x-1)\right.$
$\displaystyle \left.\qquad+15(x^3-x^2-x+1)\right]$
$\displaystyle =2(32x^3-24x)$
$\displaystyle =64x^3-48x$
$\displaystyle =16x(4x^2-3)$

$\displaystyle \text{ii) }\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6$
$\displaystyle =2\left[x^6+15x^4(x^2-1)+15x^2(x^2-1)^2+(x^2-1)^3\right]$
$\displaystyle =2\left[x^6+15x^6-15x^4+15x^6-30x^4+15x^2\right.$
$\displaystyle \left.\qquad+x^6-3x^4+3x^2-1\right]$
$\displaystyle =64x^6-96x^4+36x^2-2$

$\displaystyle \text{iii) }(1+2\sqrt{x})^5-(1-2\sqrt{x})^5$
$\displaystyle =2\left[{}^5C_1(2\sqrt{x})+{}^5C_3(2\sqrt{x})^3+{}^5C_5(2\sqrt{x})^5\right]$
$\displaystyle =2\left[10\sqrt{x}+80x\sqrt{x}+32x^2\sqrt{x}\right]$
$\displaystyle =20\sqrt{x}+160x\sqrt{x}+64x^2\sqrt{x}$
$\displaystyle =4\sqrt{x}(5+40x+16x^2)$

$\displaystyle \text{iv) }(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$
$\displaystyle =2\left[{}^6C_0(\sqrt{2})^6+{}^6C_2(\sqrt{2})^4+{}^6C_4(\sqrt{2})^2+{}^6C_6\right]$
$\displaystyle =2\left[8+15(4)+15(2)+1\right]$
$\displaystyle =2(99)$
$\displaystyle =198$

$\displaystyle \text{v) }(3+\sqrt{2})^5-(3-\sqrt{2})^5$
$\displaystyle =2\left[{}^5C_1(3)^4\sqrt{2}+{}^5C_3(3)^2(\sqrt{2})^3+{}^5C_5(\sqrt{2})^5\right]$
$\displaystyle =2\left[5(81)\sqrt{2}+10(9)(2\sqrt{2})+4\sqrt{2}\right]$
$\displaystyle =2\sqrt{2}(405+180+4)$
$\displaystyle =1178\sqrt{2}$

$\displaystyle \text{vi) }(2+\sqrt{3})^7+(2-\sqrt{3})^7$
$\displaystyle =2\left[{}^7C_0(2)^7+{}^7C_2(2)^5(\sqrt{3})^2\right.$
$\displaystyle \left.\qquad+{}^7C_4(2)^3(\sqrt{3})^4+{}^7C_6(2)(\sqrt{3})^6\right]$
$\displaystyle =2\left[128+21(32)(3)+35(8)(9)+7(2)(27)\right]$
$\displaystyle =2(128+2016+2520+378)$
$\displaystyle =10084$

$\displaystyle \text{vii) }(\sqrt{3}+1)^5-(\sqrt{3}-1)^5$
$\displaystyle =2\left[{}^5C_1(\sqrt{3})^4+{}^5C_3(\sqrt{3})^2+{}^5C_5\right]$
$\displaystyle =2\left[5(9)+10(3)+1\right]$
$\displaystyle =2(76)$
$\displaystyle =152$

$\displaystyle \text{viii) }(0.99)^5+(1.01)^5$
$\displaystyle =(1-0.01)^5+(1+0.01)^5$
$\displaystyle =2\left[1+{}^5C_2(0.01)^2+{}^5C_4(0.01)^4\right]$
$\displaystyle =2\left[1+10(0.0001)+5(0.00000001)\right]$
$\displaystyle =2(1.00100005)$
$\displaystyle =2.0020001$

$\displaystyle \text{ix) }(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6$
$\displaystyle =2\left[{}^6C_1(\sqrt{3})^5\sqrt{2}+{}^6C_3(\sqrt{3})^3(\sqrt{2})^3\right.$
$\displaystyle \left.\qquad+{}^6C_5\sqrt{3}(\sqrt{2})^5\right]$
$\displaystyle =2\left[6(9\sqrt{3})(\sqrt{2})+20(3\sqrt{3})(2\sqrt{2})\right.$
$\displaystyle \left.\qquad+6(\sqrt{3})(4\sqrt{2})\right]$
$\displaystyle =2\sqrt{6}(54+120+24)$
$\displaystyle =396\sqrt{6}$

$\displaystyle \text{x) }\left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4$
$\displaystyle =2\left[(a^2)^4+{}^4C_2(a^2)^2(a^2-1)+(a^2-1)^2\right]$
$\displaystyle =2\left[a^8+6a^4(a^2-1)+(a^2-1)^2\right]$
$\displaystyle =2\left[a^8+6a^6-6a^4+a^4-2a^2+1\right]$
$\displaystyle =2a^8+12a^6-10a^4-4a^2+2$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find }(a+b)^4-(a-b)^4.\text{ Hence, evaluate }(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4.$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b)^4-(a-b)^4$
$\displaystyle =2\left[{}^4C_1a^3b+{}^4C_3ab^3\right]$
$\displaystyle =2(4a^3b+4ab^3)$
$\displaystyle =8ab(a^2+b^2)$
$\displaystyle \text{Now, let }a=\sqrt{3}\text{ and }b=\sqrt{2}.$
$\displaystyle (\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4$
$\displaystyle =8(\sqrt{3})(\sqrt{2})\left[(\sqrt{3})^2+(\sqrt{2})^2\right]$
$\displaystyle =8\sqrt{6}(3+2)$
$\displaystyle =40\sqrt{6}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find }(x+1)^6+(x-1)^6.\text{ Hence, evaluate }(\sqrt{2}+1)^6+(\sqrt{2}-1)^6.$
$\displaystyle \text{Answer:}$
$\displaystyle (x+1)^6+(x-1)^6$
$\displaystyle =2\left[{}^6C_0x^6+{}^6C_2x^4+{}^6C_4x^2+{}^6C_6\right]$
$\displaystyle =2(x^6+15x^4+15x^2+1)$
$\displaystyle \text{Now, let }x=\sqrt{2}.$
$\displaystyle (\sqrt{2}+1)^6+(\sqrt{2}-1)^6$
$\displaystyle =2\left[(\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1\right]$
$\displaystyle =2(8+60+30+1)$
$\displaystyle =198$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Using the binomial theorem, evaluate each of the following:}$
$\displaystyle \text{i) }(96)^3\qquad\text{ii) }(102)^5\qquad\text{iii) }(101)^4\qquad\text{iv) }(98)^5$
$\displaystyle \text{Answer:}$

$\displaystyle \text{i) }(96)^3=(100-4)^3$
$\displaystyle ={}^3C_0(100)^3(-4)^0+{}^3C_1(100)^2(-4)^1+{}^3C_2(100)(-4)^2+{}^3C_3(-4)^3$
$\displaystyle =100^3-3(100)^2(4)+3(100)(4)^2-4^3$
$\displaystyle =1000000-120000+4800-64$
$\displaystyle =884736$

$\displaystyle \text{ii) }(102)^5=(100+2)^5$
$\displaystyle ={}^5C_0(100)^5+{}^5C_1(100)^4(2)+{}^5C_2(100)^3(2)^2$
$\displaystyle \qquad+{}^5C_3(100)^2(2)^3+{}^5C_4(100)(2)^4+{}^5C_5(2)^5$
$\displaystyle =100^5+5(100)^4(2)+10(100)^3(2)^2$
$\displaystyle \qquad+10(100)^2(2)^3+5(100)(2)^4+2^5$
$\displaystyle =10000000000+1000000000+40000000+800000+8000+32$
$\displaystyle =11040808032$

$\displaystyle \text{iii) }(101)^4=(100+1)^4$
$\displaystyle ={}^4C_0(100)^4(1)^0+{}^4C_1(100)^3(1)^1+{}^4C_2(100)^2(1)^2$
$\displaystyle \qquad+{}^4C_3(100)(1)^3+{}^4C_4(1)^4$
$\displaystyle =100^4+4(100)^3+6(100)^2+4(100)+1$
$\displaystyle =100000000+4000000+60000+400+1$
$\displaystyle =104060401$

$\displaystyle \text{iv) }(98)^5=(100-2)^5$
$\displaystyle ={}^5C_0(100)^5(-2)^0+{}^5C_1(100)^4(-2)^1+{}^5C_2(100)^3(-2)^2$
$\displaystyle \qquad+{}^5C_3(100)^2(-2)^3+{}^5C_4(100)(-2)^4+{}^5C_5(-2)^5$
$\displaystyle =100^5-10(100)^4+40(100)^3-80(100)^2+80(100)-32$
$\displaystyle =10000000000-1000000000+40000000-800000+8000-32$
$\displaystyle =9039207968$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Using the binomial theorem, prove that }2^{3n}-7n-1\text{ is divisible by }$
$\displaystyle 49,\text{ where }n\in N.$
$\displaystyle \text{Answer:}$
$\displaystyle 2^{3n}-7n-1$
$\displaystyle =8^n-7n-1$
$\displaystyle =(1+7)^n-7n-1$
$\displaystyle =\left[{}^nC_0+{}^nC_1(7)+{}^nC_2(7)^2+\cdots+{}^nC_n(7)^n\right]-7n-1$
$\displaystyle =1+7n+{}^nC_2(7)^2+\cdots+{}^nC_n(7)^n-7n-1$
$\displaystyle ={}^nC_2(7)^2+{}^nC_3(7)^3+\cdots+{}^nC_n(7)^n$
$\displaystyle =49\left[{}^nC_2+{}^nC_3(7)+\cdots+{}^nC_n(7)^{n-2}\right]$
$\displaystyle \text{Thus, for }n\geq2,\text{ the expression contains }49\text{ as a factor.}$
$\displaystyle \text{Also, for }n=1,\quad 2^3-7-1=0,\text{ which is divisible by }49.$
$\displaystyle \therefore 2^{3n}-7n-1\text{ is divisible by }49\text{ for all }n\in N.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Using the binomial theorem, prove that }3^{2n+2}-8n-9\text{ is divisible by }$
$\displaystyle 64,\text{ where }n\in N.$
$\displaystyle \text{Answer:}$
$\displaystyle 3^{2n+2}-8n-9$
$\displaystyle =9^{n+1}-8n-9$
$\displaystyle =(1+8)^{n+1}-8n-9$
$\displaystyle =\left[{}^{n+1}C_0+{}^{n+1}C_1(8)+{}^{n+1}C_2(8)^2+\cdots\right.$
$\displaystyle \left.\qquad+{}^{n+1}C_{n+1}(8)^{n+1}\right]-8n-9$
$\displaystyle =1+8(n+1)+{}^{n+1}C_2(8)^2+\cdots$
$\displaystyle \qquad+{}^{n+1}C_{n+1}(8)^{n+1}-8n-9$
$\displaystyle ={}^{n+1}C_2(8)^2+{}^{n+1}C_3(8)^3+\cdots+{}^{n+1}C_{n+1}(8)^{n+1}$
$\displaystyle =64\left[{}^{n+1}C_2+{}^{n+1}C_3(8)+\cdots+{}^{n+1}C_{n+1}(8)^{n-1}\right]$
$\displaystyle \therefore 3^{2n+2}-8n-9\text{ is divisible by }64\text{ for all }n\in N.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }n\text{ is a positive integer, prove that }3^{3n}-26n-1\text{ is divisible by}$
$\displaystyle 676.$
$\displaystyle \text{Answer:}$
$\displaystyle 3^{3n}-26n-1$
$\displaystyle =27^n-26n-1$
$\displaystyle =(1+26)^n-26n-1$
$\displaystyle =\left[{}^nC_0+{}^nC_1(26)+{}^nC_2(26)^2+\cdots+{}^nC_n(26)^n\right]-26n-1$
$\displaystyle =1+26n+{}^nC_2(26)^2+\cdots+{}^nC_n(26)^n-26n-1$
$\displaystyle ={}^nC_2(26)^2+{}^nC_3(26)^3+\cdots+{}^nC_n(26)^n$
$\displaystyle =676\left[{}^nC_2+{}^nC_3(26)+\cdots+{}^nC_n(26)^{n-2}\right]$
$\displaystyle \text{Thus, for }n\geq2,\text{ the expression contains }676\text{ as a factor.}$
$\displaystyle \text{Also, for }n=1,\quad 3^3-26-1=0,\text{ which is divisible by }676.$
$\displaystyle \therefore 3^{3n}-26n-1\text{ is divisible by }676\text{ for every positive integer }n.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Using the binomial theorem, indicate which is larger: }(1.1)^{10000}\text{ or }10000.$
$\displaystyle \text{Answer:}$
$\displaystyle (1.1)^{10000}=(1+0.1)^{10000}$
$\displaystyle ={}^{10000}C_0+{}^{10000}C_1(0.1)+{}^{10000}C_2(0.1)^2+\cdots$
$\displaystyle =1+10000(0.1)+\frac{10000\times9999}{2}(0.1)^2+\text{other positive terms}$
$\displaystyle =1+1000+499950+\text{other positive terms}$
$\displaystyle >500951$
$\displaystyle >10000$
$\displaystyle \therefore (1.1)^{10000}\text{ is larger than }10000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Using the binomial theorem, determine which is smaller: }(1.2)^{4000}\text{ or }800.$
$\displaystyle \text{Answer:}$
$\displaystyle (1.2)^{4000}=(1+0.2)^{4000}$
$\displaystyle ={}^{4000}C_0+{}^{4000}C_1(0.2)+{}^{4000}C_2(0.2)^2+\cdots$
$\displaystyle \qquad+{}^{4000}C_{4000}(0.2)^{4000}$
$\displaystyle =1+4000(0.2)+\text{other positive terms}$
$\displaystyle =801+\text{other positive terms}$
$\displaystyle >801$
$\displaystyle >800$
$\displaystyle \therefore 800\text{ is smaller than }(1.2)^{4000}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Find the value of }(1.01)^{10}+(1-0.01)^{10}\text{ correct to }7\text{ decimal places.}$
$\displaystyle \text{Answer:}$
$\displaystyle (1.01)^{10}+(1-0.01)^{10}$
$\displaystyle =(1+0.01)^{10}+(1-0.01)^{10}$
$\displaystyle =2\left[{}^{10}C_0+{}^{10}C_2(0.01)^2+{}^{10}C_4(0.01)^4\right.$
$\displaystyle \left.\qquad+{}^{10}C_6(0.01)^6+{}^{10}C_8(0.01)^8+(0.01)^{10}\right]$
$\displaystyle =2\left[1+45(0.0001)+210(0.00000001)+\text{terms of order }10^{-10}\right]$
$\displaystyle =2\left[1+0.0045+0.0000021+\text{terms of order }10^{-10}\right]$
$\displaystyle =2.0090042+\text{terms of order }10^{-10}$
$\displaystyle \therefore (1.01)^{10}+(1-0.01)^{10}=2.0090042\text{ correct to }7\text{ decimal places.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Show that }2^{4n+4}-15n-16,\text{ where }n\in N,\text{ is divisible by }225.$
$\displaystyle \text{Answer:}$
$\displaystyle 2^{4n+4}-15n-16$
$\displaystyle =16^{n+1}-15n-16$
$\displaystyle =(1+15)^{n+1}-15n-16$
$\displaystyle =\left[{}^{n+1}C_0+{}^{n+1}C_1(15)+{}^{n+1}C_2(15)^2+\cdots\right.$
$\displaystyle \left.\qquad+{}^{n+1}C_{n+1}(15)^{n+1}\right]-15n-16$
$\displaystyle =1+15(n+1)+{}^{n+1}C_2(15)^2+\cdots$
$\displaystyle \qquad+{}^{n+1}C_{n+1}(15)^{n+1}-15n-16$
$\displaystyle ={}^{n+1}C_2(15)^2+{}^{n+1}C_3(15)^3+\cdots$
$\displaystyle \qquad+{}^{n+1}C_{n+1}(15)^{n+1}$
$\displaystyle =225\left[{}^{n+1}C_2+{}^{n+1}C_3(15)+\cdots\right.$
$\displaystyle \left.\qquad+{}^{n+1}C_{n+1}(15)^{n-1}\right]$
$\displaystyle \therefore 2^{4n+4}-15n-16\text{ is divisible by }225\text{ for all }n\in N.$
$\displaystyle \\$