Homework #13:

10.2

	sqrt: square root
4>
	a_n = a_(n-1) + a_(n-2), n>=2, a_0=a_1=1
	r^2-r-1=0, r=(1+/-sqrt(5))/2
	a_n=A((1+sqrt(5))/2)^n+B((1-sqrt(5))/2)^n
	a0 = a1 = 1 => A = (1+sqrt(5))/(2sqrt(5)), B=(sqrt(5)-1)/(2sqrt(5))
	=>a_n

8> a_n: the number of ways to arrange n such chips with no consecutive blue chips
     b_n: the number of arrangements counted in a_n that end in blue; c_n=a_n-b_n
     Then a_(n+1) = 3b_n+4c_n=3(b_n+c_n)+c_n=3a_n+3a_n-1
     => a_n+1-3a_n-3a_(n-1)=0, n>=1, a_0=1, a_1=4
     This recurrent relation has characteristic roots r = (3+/-sqrt(21))/2
     Solve for a_n.
     a_n=[(5+sqrt(21))/(2sqrt(21))][(3+sqrt(21))/2]^n
     -[(5-sqrt(21))/(2sqrt(21))][(3-sqrt(21))/2]^n, n>=0

14> Expanding by row 1, D_n=2D_n-1 - D where D is an (n-1)by(n-1)
        determinant whose value, upon expansion by 
       its first column, is D_n-1 => D_n=2D_n-1 - D_n-2
	=> the characteristic roots r=1, 1 so D_n=A(1)^n + Bn(1)^n=A+Bn

	D_1=2,  D_2=2*2-1=3
	2=D_1=A+B; 3=D_2=A+2B=>A=B=1 and D_n=1+n, n>=1

16>     b=-8, c=7

10.6
8>	a> use the Alternative Form of the Principle of Mathematical Induction
	b> O(n)