$f(x) =1-x+x^2-x^3+\cdots+x^{16}-x^{17}$

$=a_0+a_1(1+x)+a_2(1+x)^2+ a_3(1+x)^3+\ldots+a_{17}(1+x)^{17}$

What is $a_2$ ?

$1=a_0+a_1+a_2+\ldots+a_{17}$

$-x=a_1 x+2 a_2 x+3 a_3 x+\cdots+17 a_{17} x$

$x^2=a_2 x^2+{ }^3 c_2^{a_3 x^2}+{ }^4 c_2^{a_4 x^2} +{ }^5 c_2^{a^5 x^2}+\cdots \cdot{ }^{17} c_2 a_{17} x^2$

$\frac{3 !}{12!1!} \frac{4}{2} \frac{4 !}{2 ! 2 !} \frac{5}{3} \frac{5 !}{2 ! 3 !} \rightarrow \frac{6}{4}$

$f^\prime=\frac{d f}{d x}$

$=-1+2 x-3 x^2+\cdots+16 x^{15}-17 x^{16}$

$=a_1+2 a_2(1+x)+3 a_3(1+x)^2 +4 a_4(1+x)^3+\ldots+17 a_{17}(1+x)^{16}$

$f^{\prime \prime}=\frac{d^2 f}{d x^2}$

$= 2-6 x+4 \times 3 x^2-5 \times 4 x^3$

$+\cdots+16 \times 15 x^{14}-17 \times 16 x^{15}$

$= 2a_2+3 a_3 \times 2(1+x)+4 a_4 \times 3(1+x)^2$

$+\cdots+17 a_{17} \times 16(1+x)^{15}$

$2 a_2=2 \times+3 \times2+4 \times 3+5 \times 4+6 \times 5+\cdots+17 \times 16 \quad x=-1$

$2 a_2=2 \times+3 \times2+4 \times 3+5 \times 4+6 \times 5+\cdots+17 \times 16$

$x=-1$

$=2+6+12+20+30+42+\cdots+17 \times 16$

$=\sum_{n=2}^{17} n(n-1)=\sum_{n=2}^{17} n^2-\sum_{n=2}^{17} n$

$\biggl \{ \because (1+2+3+\ldots+n) + (n+(n-1)+(n-3+\ldots+1) = n(n+1) \\ 1^2+2^2+3^2+\cdots+n^2 = \frac{n (n+1) (2n+1)}{6} \biggl \}$

$\therefore \sum_{n=2}^{17} n^2-\sum_{n=2}^{17} n = \left(\frac{7 \times 18 \times 35}{6}-1\right)-\left(\frac{17 \times 18}{2}-1\right)$

$=17 \times 18^3 \times\left(\frac{32}{6}\right)=51 \times 32$

$a_2=51 \times 16 =816$

$(3+\sqrt{5})^n =I+f$ where $I$ is the largest integer and $0 \leq f < 1$

Solution: Suppose $(3+\sqrt{5})^n =\rho+\sigma$

$\rho$ = rational and $\sigma$ = irrational

We need to show that
$\rho = \frac{I+2}{2}, \ \ \sigma = \frac{I+2f-1}{2}$

$(3+\sqrt{5})^n \longrightarrow 3^n + C_ 2 3^{n-2} \cdot 5 + … C_{n-1} 3 \cdot 5^{ \frac{n}{2}}$

$(3-\sqrt{5})^n \longrightarrow C_ 1 3^{n-1} \sqrt{5} + C_ 3 3^{n-3} \cdot 5 \cdot 15 + … C_n \cdot 5^{\frac{n}{2}} \cdot \sqrt{5}$

$(3+\sqrt5)^n=\rho+\sigma=I+f$

$(3-\sqrt5)^n=\rho-\sigma$

$(3+\sqrt5)^n=I+f$

$(3-\sqrt5)^n=(I_{2^{-1}})+(1-f)$

$0<(3-\sqrt{5})^n<1$

$(7+4 \sqrt{3})^n=I+f$

Show that

$(1-f)(I+f)=1$

Solution:

$(7-4 \sqrt{3})^n- =1-f$

$(1-f)(I+f) =(7-4 \sqrt{3})^n(7+4 \sqrt{3})^n$

$=[(7-4 \sqrt{3})(7+4 \sqrt{3})]^n$

$=[49-48]^n=1 .$

$P=(2+\sqrt{3})^5, f$ is the fractional part of $P$.

$f=P-[P]$

Find $\frac{f^2}{1-f}$

Solution:

$(2-\sqrt{3})^5 \leftarrow \text { fractional term }=1-f$

$(2-\sqrt{3})^5=1-f$

$f=1-(2-\sqrt{3})^5$

$\frac{f^2}{1-f}= \frac{[1-(2-\sqrt{3})^5]^2}{(2-\sqrt{3})^5}$

$=\frac{(2+\sqrt{3})^5\left(1-(2-\sqrt{3})^5\right)^2}{[(2+\sqrt{3})(2-\sqrt{3})]^5}$

$=(2+\sqrt{3})^5-2(2+\sqrt{3})^5(2-\sqrt{3})^5 +(2+\sqrt{3})^5(2-\sqrt{3})^{10}$

$\frac{f^2}{1-f} =[(2+\sqrt{3})^5+(2-\sqrt{3})^5]-2$

$=2 \times\left(2^5+2^3 \cdot 10 \cdot 3+2 \cdot 5 \cdot 9\right)-2$

$=2 \times(32+240+90)-2$

$=2 \times 362-2=722 .$

Find the sum of the roots of the equation:

$x^{2001}+\left(\frac{1}{2}-x\right)^{2001}=0$

Solution: $P_1, P_2, P_3 \ldots P_{2000}$

$(\frac{1}{2} )\left(x-P_1\right)\left(x-P_2\right)\left(x-P_3\right) \cdots$

$\left(x-P_{2000}\right)=0$

$x^{2001}+\left(\frac{1}{2}-x\right)^{2001} = \frac{1}{2}\left(x-p_1\right)\left(x-P_2\right) \cdots\left(x-P_{2000}\right)$

$x^{2000}-x^{1900} \sum_{i=1}^{2000} P_i+\cdots$

$^{2001} C_2 x^{1999}$

$\sum P_i={ }^{2001} C_2=\frac{2001 \times 2000}{2}$