mathematics-class-11-unit-09-chapter-05-binomial-theorem-l-8-9-a0skbsb29mu

### <Success>Problem</Success>
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<p>$f(x) =1-x+x^2-x^3+\cdots+x^{16}-x^{17}$</p>
<p>$=a_0+a_1(1+x)+a_2(1+x)^2+ a_3(1+x)^3+\ldots+a_{17}(1+x)^{17}$</p>
<p>What is $a_2$ ?</p>
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<p>======</p>
<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>$1=a_0+a_1+a_2+\ldots+a_{17}$</p>
<p>$-x=a_1 x+2 a_2 x+3 a_3 x+\cdots+17 a_{17} x$</p>
<p>$x^2=a_2 x^2+{ }^3 c_2^{a_3 x^2}+{ }^4 c_2^{a_4 x^2} +{ }^5 c_2^{a^5 x^2}+\cdots \cdot{ }^{17} c_2 a_{17} x^2$</p>
<p>$\frac{3 !}{12!1!} \frac{4}{2} \frac{4 !}{2 ! 2 !} \frac{5}{3} \frac{5 !}{2 ! 3 !} \rightarrow \frac{6}{4}$</p>
<p>$f^\prime=\frac{d f}{d x}$</p>
<p>$=-1+2 x-3 x^2+\cdots+16 x^{15}-17 x^{16}$</p>
<p>$=a_1+2 a_2(1+x)+3 a_3(1+x)^2 +4 a_4(1+x)^3+\ldots+17 a_{17}(1+x)^{16}$</p>
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### <Success>Solution</Success>
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<p>$2 a_2=2 \times+3 \times2+4 \times 3+5 \times 4+6 \times 5+\cdots+17 \times 16 $</p>
<p>$ x=-1$</p>
<p>$=2+6+12+20+30+42+\cdots+17 \times 16$</p>
<p>$=\sum_{n=2}^{17} n(n-1)=\sum_{n=2}^{17} n^2-\sum_{n=2}^{17} n$</p>
<p>$\biggl \{ \because (1+2+3+\ldots+n) + (n+(n-1)+(n-3+\ldots+1) = n(n+1) \\ 1^2+2^2+3^2+\cdots+n^2 = \frac{n (n+1) (2n+1)}{6} \biggl \}$</p>
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<p>======</p>
<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
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<p>$\therefore \sum_{n=2}^{17} n^2-\sum_{n=2}^{17} n = \left(\frac{7 \times 18 \times 35}{6}-1\right)-\left(\frac{17 \times 18}{2}-1\right)$</p>
<p>$=17 \times 18^3 \times\left(\frac{32}{6}\right)=51 \times 32 $</p>
<p>$a_2=51 \times 16  =816 $</p>
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### <Success>Problem</Success>
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<p>$(3+\sqrt{5})^n  =I+f$ where $I$ is the largest integer and $0 \leq f < 1 $</p>
<p><strong>Solution:</strong>   Suppose $(3+\sqrt{5})^n  =\rho+\sigma $</p>
<p>$\rho$ = rational and $\sigma$ = irrational</p>
<p>We need to show that<br>
$ \rho = \frac{I+2}{2}, \ \ \sigma = \frac{I+2f-1}{2}$</p>
<p>$(3+\sqrt{5})^n \longrightarrow 3^n + C_ 2 3^{n-2} \cdot 5 + … C_{n-1} 3 \cdot 5^{ \frac{n}{2}} $</p>
<p>$(3-\sqrt{5})^n \longrightarrow C_ 1 3^{n-1} \sqrt{5} + C_ 3 3^{n-3} \cdot 5 \cdot 15 + … C_n \cdot 5^{\frac{n}{2}} \cdot \sqrt{5}$</p>
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<p>======</p>
<h3 id="successsolutionsuccess-2"><Success>Solution</Success></h3>
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<p>$ (3+\sqrt5)^n=\rho+\sigma=I+f $</p>
<p>$ (3-\sqrt5)^n=\rho-\sigma$</p>
<p>$(3+\sqrt5)^n=I+f $</p>
<p>$(3-\sqrt5)^n=(I_{2^{-1}})+(1-f)$</p>
<p>$0<(3-\sqrt{5})^n<1$</p>
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### <Success>Problem</Success>
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<p>$P=(2+\sqrt{3})^5, f$ is the fractional part of $P$.</p>
<p>$f=P-[P]$</p>
<p>Find $\frac{f^2}{1-f}$</p>
<p><strong>Solution:</strong></p>
<p>$(2-\sqrt{3})^5 \leftarrow \text { fractional term }=1-f$</p>
<p>$(2-\sqrt{3})^5=1-f$</p>
<p>$f=1-(2-\sqrt{3})^5$</p>
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### <Success>Solution</Success>
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<p>$\frac{f^2}{1-f}= \frac{[1-(2-\sqrt{3})^5]^2}{(2-\sqrt{3})^5}$</p>
<p>$=\frac{(2+\sqrt{3})^5\left(1-(2-\sqrt{3})^5\right)^2}{[(2+\sqrt{3})(2-\sqrt{3})]^5}$</p>
<p>$=(2+\sqrt{3})^5-2(2+\sqrt{3})^5(2-\sqrt{3})^5 +(2+\sqrt{3})^5(2-\sqrt{3})^{10}$</p>
<p>$\frac{f^2}{1-f} =[(2+\sqrt{3})^5+(2-\sqrt{3})^5]-2$</p>
<p>$ =2 \times\left(2^5+2^3 \cdot 10 \cdot 3+2 \cdot 5 \cdot 9\right)-2$</p>
<p>$=2 \times(32+240+90)-2$</p>
<p>$=2 \times 362-2=722 .$</p>
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### <Success>Problem</Success>
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<p>Find the sum of the roots of the equation:</p>
<p>$x^{2001}+\left(\frac{1}{2}-x\right)^{2001}=0$</p>
<p><strong>Solution:</strong> $ P_1, P_2, P_3 \ldots P_{2000}$</p>
<p>$(\frac{1}{2} )\left(x-P_1\right)\left(x-P_2\right)\left(x-P_3\right) \cdots $</p>
<p>$\left(x-P_{2000}\right)=0$</p>
<p>$ x^{2001}+\left(\frac{1}{2}-x\right)^{2001} = \frac{1}{2}\left(x-p_1\right)\left(x-P_2\right) \cdots\left(x-P_{2000}\right)$</p>
<p>$ x^{2000}-x^{1900} \sum_{i=1}^{2000} P_i+\cdots$</p>
<p>$^{2001} C_2 x^{1999}$</p>
<p>$\sum P_i={ }^{2001} C_2=\frac{2001 \times 2000}{2}$</p>
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<p>======</p>
<h3 id="successthank-yousuccess"><Success>Thank you</Success></h3>
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