Answer :
### Question 22
We are given that:
1. The 6th term of the sequence is [tex]\( t_6 = 50 \)[/tex].
2. The recurrence relation for the sequence is given by [tex]\( t_n = 2t_{n-1} + 5n^2 + n \)[/tex].
Our goal is to find the first five terms [tex]\( t_1, t_2, t_3, t_4, t_5 \)[/tex].
Let's start by working backwards from [tex]\( t_6 \)[/tex] to find each previous term.
1. Finding [tex]\( t_5 \)[/tex]:
Given [tex]\( t_6 = 50 \)[/tex],
[tex]\[ t_6 = 2t_5 + 5 \cdot 6^2 + 6 \][/tex]
[tex]\[ 50 = 2t_5 + 5 \cdot 36 + 6 \][/tex]
[tex]\[ 50 = 2t_5 + 180 + 6 \][/tex]
[tex]\[ 50 = 2t_5 + 186 \][/tex]
[tex]\[ 2t_5 = 50 - 186 \][/tex]
[tex]\[ 2t_5 = -136 \][/tex]
[tex]\[ t_5 = -68 \][/tex]
2. Finding [tex]\( t_4 \)[/tex]:
[tex]\[ t_5 = 2t_4 + 5 \cdot 5^2 + 5 \][/tex]
[tex]\[ -68 = 2t_4 + 5 \cdot 25 + 5 \][/tex]
[tex]\[ -68 = 2t_4 + 125 + 5 \][/tex]
[tex]\[ -68 = 2t_4 + 130 \][/tex]
[tex]\[ 2t_4 = -68 - 130 \][/tex]
[tex]\[ 2t_4 = -198 \][/tex]
[tex]\[ t_4 = -99 \][/tex]
3. Finding [tex]\( t_3 \)[/tex]:
[tex]\[ t_4 = 2t_3 + 5 \cdot 4^2 + 4 \][/tex]
[tex]\[ -99 = 2t_3 + 5 \cdot 16 + 4 \][/tex]
[tex]\[ -99 = 2t_3 + 80 + 4 \][/tex]
[tex]\[ -99 = 2t_3 + 84 \][/tex]
[tex]\[ 2t_3 = -99 - 84 \][/tex]
[tex]\[ 2t_3 = -183 \][/tex]
[tex]\[ t_3 = -91.5 \][/tex]
4. Finding [tex]\( t_2 \)[/tex]:
[tex]\[ t_3 = 2t_2 + 5 \cdot 3^2 + 3 \][/tex]
[tex]\[ -91.5 = 2t_2 + 5 \cdot 9 + 3 \][/tex]
[tex]\[ -91.5 = 2t_2 + 45 + 3 \][/tex]
[tex]\[ -91.5 = 2t_2 + 48 \][/tex]
[tex]\[ 2t_2 = -91.5 - 48 \][/tex]
[tex]\[ 2t_2 = -139.5 \][/tex]
[tex]\[ t_2 = -69.75 \][/tex]
5. Finding [tex]\( t_1 \)[/tex]:
[tex]\[ t_2 = 2t_1 + 5 \cdot 2^2 + 2 \][/tex]
[tex]\[ -69.75 = 2t_1 + 5 \cdot 4 + 2 \][/tex]
[tex]\[ -69.75 = 2t_1 + 20 + 2 \][/tex]
[tex]\[ -69.75 = 2t_1 + 22 \][/tex]
[tex]\[ 2t_1 = -69.75 - 22 \][/tex]
[tex]\[ 2t_1 = -91.75 \][/tex]
[tex]\[ t_1 = -45.875 \][/tex]
So, the first five terms of the sequence are:
[tex]\[ t_1 = -45.875, \][/tex]
[tex]\[ t_2 = -69.75, \][/tex]
[tex]\[ t_3 = -91.5, \][/tex]
[tex]\[ t_4 = -99, \][/tex]
[tex]\[ t_5 = -68. \][/tex]
### Question 23
Given the function [tex]\( f \)[/tex] satisfying the property [tex]\( f(x+3) = f(x) + f(3) \)[/tex] for all [tex]\( x \in \mathbb{R} \)[/tex], we need to prove that [tex]\( f(0) = 0 \)[/tex] and [tex]\( f(-3) = -f(3) \)[/tex].
1. Proving [tex]\( f(0) = 0 \)[/tex]:
Let [tex]\( x = 0 \)[/tex] in the given functional equation:
[tex]\[ f(0+3) = f(0) + f(3) \][/tex]
[tex]\[ f(3) = f(0) + f(3) \][/tex]
Subtracting [tex]\( f(3) \)[/tex] from both sides:
[tex]\[ 0 = f(0) \][/tex]
Hence, [tex]\( f(0) = 0 \)[/tex].
2. Proving [tex]\( f(-3) = -f(3) \)[/tex]:
Now let [tex]\( x = -3 \)[/tex] in the given functional equation:
[tex]\[ f(-3+3) = f(-3) + f(3) \][/tex]
[tex]\[ f(0) = f(-3) + f(3) \][/tex]
We already know that [tex]\( f(0) = 0 \)[/tex], so:
[tex]\[ 0 = f(-3) + f(3) \][/tex]
Rearranging this equation:
[tex]\[ f(-3) = -f(3) \][/tex]
Therefore, we have proven that [tex]\( f(0) = 0 \)[/tex] and [tex]\( f(-3) = -f(3) \)[/tex].
We are given that:
1. The 6th term of the sequence is [tex]\( t_6 = 50 \)[/tex].
2. The recurrence relation for the sequence is given by [tex]\( t_n = 2t_{n-1} + 5n^2 + n \)[/tex].
Our goal is to find the first five terms [tex]\( t_1, t_2, t_3, t_4, t_5 \)[/tex].
Let's start by working backwards from [tex]\( t_6 \)[/tex] to find each previous term.
1. Finding [tex]\( t_5 \)[/tex]:
Given [tex]\( t_6 = 50 \)[/tex],
[tex]\[ t_6 = 2t_5 + 5 \cdot 6^2 + 6 \][/tex]
[tex]\[ 50 = 2t_5 + 5 \cdot 36 + 6 \][/tex]
[tex]\[ 50 = 2t_5 + 180 + 6 \][/tex]
[tex]\[ 50 = 2t_5 + 186 \][/tex]
[tex]\[ 2t_5 = 50 - 186 \][/tex]
[tex]\[ 2t_5 = -136 \][/tex]
[tex]\[ t_5 = -68 \][/tex]
2. Finding [tex]\( t_4 \)[/tex]:
[tex]\[ t_5 = 2t_4 + 5 \cdot 5^2 + 5 \][/tex]
[tex]\[ -68 = 2t_4 + 5 \cdot 25 + 5 \][/tex]
[tex]\[ -68 = 2t_4 + 125 + 5 \][/tex]
[tex]\[ -68 = 2t_4 + 130 \][/tex]
[tex]\[ 2t_4 = -68 - 130 \][/tex]
[tex]\[ 2t_4 = -198 \][/tex]
[tex]\[ t_4 = -99 \][/tex]
3. Finding [tex]\( t_3 \)[/tex]:
[tex]\[ t_4 = 2t_3 + 5 \cdot 4^2 + 4 \][/tex]
[tex]\[ -99 = 2t_3 + 5 \cdot 16 + 4 \][/tex]
[tex]\[ -99 = 2t_3 + 80 + 4 \][/tex]
[tex]\[ -99 = 2t_3 + 84 \][/tex]
[tex]\[ 2t_3 = -99 - 84 \][/tex]
[tex]\[ 2t_3 = -183 \][/tex]
[tex]\[ t_3 = -91.5 \][/tex]
4. Finding [tex]\( t_2 \)[/tex]:
[tex]\[ t_3 = 2t_2 + 5 \cdot 3^2 + 3 \][/tex]
[tex]\[ -91.5 = 2t_2 + 5 \cdot 9 + 3 \][/tex]
[tex]\[ -91.5 = 2t_2 + 45 + 3 \][/tex]
[tex]\[ -91.5 = 2t_2 + 48 \][/tex]
[tex]\[ 2t_2 = -91.5 - 48 \][/tex]
[tex]\[ 2t_2 = -139.5 \][/tex]
[tex]\[ t_2 = -69.75 \][/tex]
5. Finding [tex]\( t_1 \)[/tex]:
[tex]\[ t_2 = 2t_1 + 5 \cdot 2^2 + 2 \][/tex]
[tex]\[ -69.75 = 2t_1 + 5 \cdot 4 + 2 \][/tex]
[tex]\[ -69.75 = 2t_1 + 20 + 2 \][/tex]
[tex]\[ -69.75 = 2t_1 + 22 \][/tex]
[tex]\[ 2t_1 = -69.75 - 22 \][/tex]
[tex]\[ 2t_1 = -91.75 \][/tex]
[tex]\[ t_1 = -45.875 \][/tex]
So, the first five terms of the sequence are:
[tex]\[ t_1 = -45.875, \][/tex]
[tex]\[ t_2 = -69.75, \][/tex]
[tex]\[ t_3 = -91.5, \][/tex]
[tex]\[ t_4 = -99, \][/tex]
[tex]\[ t_5 = -68. \][/tex]
### Question 23
Given the function [tex]\( f \)[/tex] satisfying the property [tex]\( f(x+3) = f(x) + f(3) \)[/tex] for all [tex]\( x \in \mathbb{R} \)[/tex], we need to prove that [tex]\( f(0) = 0 \)[/tex] and [tex]\( f(-3) = -f(3) \)[/tex].
1. Proving [tex]\( f(0) = 0 \)[/tex]:
Let [tex]\( x = 0 \)[/tex] in the given functional equation:
[tex]\[ f(0+3) = f(0) + f(3) \][/tex]
[tex]\[ f(3) = f(0) + f(3) \][/tex]
Subtracting [tex]\( f(3) \)[/tex] from both sides:
[tex]\[ 0 = f(0) \][/tex]
Hence, [tex]\( f(0) = 0 \)[/tex].
2. Proving [tex]\( f(-3) = -f(3) \)[/tex]:
Now let [tex]\( x = -3 \)[/tex] in the given functional equation:
[tex]\[ f(-3+3) = f(-3) + f(3) \][/tex]
[tex]\[ f(0) = f(-3) + f(3) \][/tex]
We already know that [tex]\( f(0) = 0 \)[/tex], so:
[tex]\[ 0 = f(-3) + f(3) \][/tex]
Rearranging this equation:
[tex]\[ f(-3) = -f(3) \][/tex]
Therefore, we have proven that [tex]\( f(0) = 0 \)[/tex] and [tex]\( f(-3) = -f(3) \)[/tex].
