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To prove by induction that for all positive integers [tex]\( n \)[/tex], [tex]\( 21 \mid \left(4^{n+1} + 5^{2n-1}\right) \)[/tex], we follow these steps:
### Base Case
Let's check the base case when [tex]\( n = 1 \)[/tex]:
[tex]\[ 4^{n+1} + 5^{2n-1} = 4^{1+1} + 5^{2 \cdot 1 - 1} = 4^2 + 5^1 = 16 + 5 = 21 \][/tex]
Since [tex]\( 21 \)[/tex] is divisible by [tex]\( 21 \)[/tex]:
[tex]\[ 21 \mid 21 \][/tex]
The base case holds true.
### Inductive Step
Next, we assume that the proposition is true for some positive integer [tex]\( k \)[/tex]. That is, we assume:
[tex]\[ 21 \mid \left( 4^{k+1} + 5^{2k-1} \right) \][/tex]
This is our induction hypothesis.
We need to prove that the proposition is true for [tex]\( k+1 \)[/tex], i.e., we need to show:
[tex]\[ 21 \mid \left( 4^{(k+1)+1} + 5^{2(k+1)-1} \right) \][/tex]
Simplify the expression:
[tex]\[ 4^{(k+1)+1} + 5^{2(k+1)-1} = 4^{k+2} + 5^{2k+1} \][/tex]
We need to show that [tex]\( 4^{k+2} + 5^{2k+1} \)[/tex] is divisible by [tex]\( 21 \)[/tex].
Express [tex]\( 4^{k+2} \)[/tex] in terms of [tex]\( 4^{k+1} \)[/tex]:
[tex]\[ 4^{k+2} = 4 \cdot 4^{k+1} \][/tex]
Similarly, express [tex]\( 5^{2k+1} \)[/tex] in terms of [tex]\( 5^{2k-1} \)[/tex]:
[tex]\[ 5^{2k+1} = 25 \cdot 5^{2k-1} \][/tex]
Thus, our problem reduces to:
[tex]\[ 4^{k+2} + 5^{2k+1} = 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \][/tex]
Using our induction hypothesis [tex]\( 21 \mid \left( 4^{k+1} + 5^{2k-1} \right) \)[/tex]:
[tex]\[ 4^{k+1} + 5^{2k-1} = 21m \quad \text{for some integer } m \][/tex]
We need to find a way to express [tex]\( 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \)[/tex] in terms of [tex]\( 21 \)[/tex]. Using the fact that:
[tex]\[ 4^{k+1} \equiv -5^{2k-1} \pmod{21} \][/tex]
By substitution [tex]\( 4 \cdot 4^{k+1} \equiv 4 \cdot (-5^{2k-1}) = -4 \cdot 5^{2k-1} \pmod{21} \)[/tex]:
[tex]\[ \Rightarrow 4 \cdot 4^{k+1} \equiv 4 \cdot 4^{k+1} \pmod{21} \][/tex]
Similarly for [tex]\( 25 \cdot 5^{2k-1} \)[/tex]:
[tex]\[ 25 \equiv 4 \pmod{21}, \][/tex]
so
[tex]\[ 25 \cdot 5^{2k-1} \equiv 4 \cdot 5^{2k-1} \pmod{21}. \][/tex]
Combining these results:
[tex]\[ 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \equiv 4 \cdot 4^{k+1} + 4 \cdot 5^{2k-1} \equiv 0 \pmod{21}. \][/tex]
Therefore,
[tex]\[ 21 \mid \left( 4^{(k+1)+1} + 5^{2(k+1)-1} \right) \][/tex]
Thus, by induction, for all positive integers [tex]\( n \)[/tex]:
[tex]\[ 21 \mid \left( 4^{n+1} + 5^{2n-1} \right) \][/tex]
This concludes the proof by mathematical induction.
### Base Case
Let's check the base case when [tex]\( n = 1 \)[/tex]:
[tex]\[ 4^{n+1} + 5^{2n-1} = 4^{1+1} + 5^{2 \cdot 1 - 1} = 4^2 + 5^1 = 16 + 5 = 21 \][/tex]
Since [tex]\( 21 \)[/tex] is divisible by [tex]\( 21 \)[/tex]:
[tex]\[ 21 \mid 21 \][/tex]
The base case holds true.
### Inductive Step
Next, we assume that the proposition is true for some positive integer [tex]\( k \)[/tex]. That is, we assume:
[tex]\[ 21 \mid \left( 4^{k+1} + 5^{2k-1} \right) \][/tex]
This is our induction hypothesis.
We need to prove that the proposition is true for [tex]\( k+1 \)[/tex], i.e., we need to show:
[tex]\[ 21 \mid \left( 4^{(k+1)+1} + 5^{2(k+1)-1} \right) \][/tex]
Simplify the expression:
[tex]\[ 4^{(k+1)+1} + 5^{2(k+1)-1} = 4^{k+2} + 5^{2k+1} \][/tex]
We need to show that [tex]\( 4^{k+2} + 5^{2k+1} \)[/tex] is divisible by [tex]\( 21 \)[/tex].
Express [tex]\( 4^{k+2} \)[/tex] in terms of [tex]\( 4^{k+1} \)[/tex]:
[tex]\[ 4^{k+2} = 4 \cdot 4^{k+1} \][/tex]
Similarly, express [tex]\( 5^{2k+1} \)[/tex] in terms of [tex]\( 5^{2k-1} \)[/tex]:
[tex]\[ 5^{2k+1} = 25 \cdot 5^{2k-1} \][/tex]
Thus, our problem reduces to:
[tex]\[ 4^{k+2} + 5^{2k+1} = 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \][/tex]
Using our induction hypothesis [tex]\( 21 \mid \left( 4^{k+1} + 5^{2k-1} \right) \)[/tex]:
[tex]\[ 4^{k+1} + 5^{2k-1} = 21m \quad \text{for some integer } m \][/tex]
We need to find a way to express [tex]\( 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \)[/tex] in terms of [tex]\( 21 \)[/tex]. Using the fact that:
[tex]\[ 4^{k+1} \equiv -5^{2k-1} \pmod{21} \][/tex]
By substitution [tex]\( 4 \cdot 4^{k+1} \equiv 4 \cdot (-5^{2k-1}) = -4 \cdot 5^{2k-1} \pmod{21} \)[/tex]:
[tex]\[ \Rightarrow 4 \cdot 4^{k+1} \equiv 4 \cdot 4^{k+1} \pmod{21} \][/tex]
Similarly for [tex]\( 25 \cdot 5^{2k-1} \)[/tex]:
[tex]\[ 25 \equiv 4 \pmod{21}, \][/tex]
so
[tex]\[ 25 \cdot 5^{2k-1} \equiv 4 \cdot 5^{2k-1} \pmod{21}. \][/tex]
Combining these results:
[tex]\[ 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \equiv 4 \cdot 4^{k+1} + 4 \cdot 5^{2k-1} \equiv 0 \pmod{21}. \][/tex]
Therefore,
[tex]\[ 21 \mid \left( 4^{(k+1)+1} + 5^{2(k+1)-1} \right) \][/tex]
Thus, by induction, for all positive integers [tex]\( n \)[/tex]:
[tex]\[ 21 \mid \left( 4^{n+1} + 5^{2n-1} \right) \][/tex]
This concludes the proof by mathematical induction.
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