Number Sequences: Difficult Problems with Solutions

Solution:
The characteristic equation to the recurrence relation is
[tex]r^2-(2+\sqrt{3})r+2\sqrt{3}[/tex]. Its roots are [tex]r_1=2[/tex] and [tex]r_2=\sqrt{3}[/tex]. Therefore
[tex]a_n=c_1.2^n+c_2.\sqrt{3}^n[/tex]
[tex]\begin{array}{|l}a_1=c_1.2+c_2.\sqrt{3}=6+\sqrt{3}\\a_2=c_1.4+c_2.3=15 \end{array}[/tex]
[tex]\begin{array}{|l}c_1.4+c_2.2\sqrt{3}=12+2\sqrt{3}\\c_1.4+c_2.3=15 \end{array}[/tex], we subtract them and get
[tex]4c_1+2\sqrt{3}.c_2-4c_1-3c_2=12+2\sqrt{3}-15[/tex], or [tex](2\sqrt{3}-3).c_2=2\sqrt{3}-3[/tex], therefore [tex]c_2=1[/tex]. Then [tex]c_1=\frac{6+\sqrt{3}-c_2\sqrt{3}}{2}=\frac{6}{2}=3[/tex] and
[tex]a_n=3.2^n+\sqrt{3}^n[/tex]. Substituting with [tex]n=8[/tex], we get
[tex]a_8=3.2^8+\sqrt{3}^8=3.256+3^4=768+81=849[/tex]

Solution:
The characteristic equation of the recurrence relation is
[tex]r^2-4r+4=0[/tex], with the double root [tex]r_{1,2}=2[/tex]. Therefore [tex]a_n[/tex] has the form
[tex]a_n=c_1.2^n+c_2.n.2^n=(c_1+c_2n).2^n[/tex]. We calculate them for [tex]n=1;2[/tex]:
[tex]\begin{array}{|l}a_1=(c_1+c_2).2=3\\(c_1+2c_2).4=7 \end{array}[/tex]
[tex]\begin{array}{|l}c_1+c_2=\frac{6}{4}\\c_1+2c_2=\frac{7}{4}\end{array}[/tex]. We subtract them to get
[tex]c_2=\frac{7}{4}-\frac{6}{4}=\frac{1}{4}[/tex]. [tex]c_1=\frac{6}{4}-c_2=\frac{5}{4}[/tex] and
[tex]a_n=(\frac{1}{4}n+\frac{5}{4}).2^n=(n+5).2^{n-2}[/tex]. Substituting n for 7, we get
[tex]a_7=(7+5).2^{7-2}=12.2^5=12.32=384[/tex]

Solution:
[tex]a_{n+1}-a_n=3^{n+1}-5^{n+1}+\frac{1}{n+1}-(3^n-5^n+\frac{1}{n})=3.3^n-5.5^n+\frac{1}{n+1}-3^n+5^n-\frac{1}{n}=(2.3^n-4.5^n)+(\frac{1}{n+1}-\frac{1}{n})[/tex].
[tex]\frac{1}{n+1}-\frac{1}{n}=\frac{n-(n+1)}{n(n+1)}=-\frac{1}{n+1}<0[/tex].
[tex]2.3^n-4.5^n=2.3^n-2.5^n-2.5^n=2.(3^n-5^n)-5^n < 2.(3^n-5^n)<0[/tex], since for positive n [tex]5^n>3^n[/tex].
The sum of these two negative expressions is the difference between members of the sequence, therefore it is decreasing.

Solution:
We rewrite the recurrence relation as [tex]a_{n+2}-a_{n+1}=a_n[/tex]. If [tex]a_n[/tex] were to be positive for any n, then [tex]a_{n+2}-a_{n+1}>0[/tex] - and it is the difference between two consecutive members of the sequence. We shall prove that [tex]a_n>0[/tex] for every n by (strong) induction.
For [tex]n=1[/tex]: [tex]a_1=1[/tex] and it is positive.
For [tex]n=2[/tex]: [tex]a_2=1[/tex] and it is positive.
Let us assume that for [tex]n \le k[/tex], where k is an aribtrary positive integer, [tex]a_n[/tex] is positive. [tex]a_{k+1}=a_k+a_{k-1}[/tex] and by the induction hypothesis both [tex]a_k[/tex] and [tex]a_{k-1}[/tex] are positive. Therefore [tex]a_n>0[/tex] for any n and since [tex]a_{n+2}-a_{n+1}=a_n>0[/tex], the sequence is increasing.