Number Sequences: Very Difficult Problems with Solutions

Solution:
You’re given the recurrence:

$a_n + a_{n-1} = 2^n,\quad a_1 = 1,; a_2 = 1$

Rewrite it as:
$a_n = 2^n - a_{n-1}$

Now compute step by step:

$a_3 = 2^3 - a_2 = 8 - 1 = 7$
$a_4 = 2^4 - a_3 = 16 - 7 = 9$
$a_5 = 2^5 - a_4 = 32 - 9 = 23$
$a_6 = 2^6 - a_5 = 64 - 23 = 41$
$a_7 = 2^7 - a_6 = 128 - 41 = 87$
$a_8 = 2^8 - a_7 = 256 - 87 = 169$
$a_9 = 2^9 - a_8 = 512 - 169 = 343$
$a_{10} = 2^{10} - a_9 = 1024 - 343 = 681$
$a_{11} = 2^{11} - a_{10} = 2048 - 681 = 1367$

Final answer:
$a_{11} = 1367$

Solution:
The recurrence relation is
[tex]a_{n+1}.a_n=a_n^2+1[/tex]
[tex]a_{n+1}=\frac{1}{a_n}+a_n[/tex]
[tex]a_{n+1}-a_n=\frac{1}{a_n}[/tex].
It is enough to prove that [tex]a_n>0[/tex] for any n. We shall do this by using mathematical induction.
For [tex]n=1[/tex]: [tex]a_1=1>0[/tex]
For [tex]n=2[/tex]: [tex]a_2=a_1+\frac{1}{a_1}=1+1=2>0[/tex]
Let us assume that this is true for [tex]n=k[/tex], where k is an arbitrary positive integer. [tex]a_{k+1}=a_k+\frac{1}{a_k}[/tex], which is the sum of two positive numbers and is greater than zero. Therefore [tex]a_n > 0[/tex] for any [tex]n[/tex].
[tex]a_{n+1}-a_n=\frac{1}{a_n}>0[/tex] for any [tex]n[/tex], therefore the sequence is increasing.