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DeathKnight
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The question is: Prove by mathematical Induction that
[tex] f(n) \equiv 2^{6n}+3^{2n-2} [/tex] is divisible by 5. This is what I did:
Suppose that the given statement is true for [tex]n=k[/tex]
Since the[tex] f(k)[/tex] is divisible by 5,
[tex]f(k)=5A[/tex] (where A are is a constant.)
Also, from the given statement:
[tex] f(k)=2^{6k}+3^{2k-2} [/tex]
To prove that the given statement is also true for n=k+1:
[tex] f(k+1)-f(k) [/tex]
[tex]=2^{6k+6}+3^{2k} - (2^{6k}+3^{2k-2})[/tex]
[tex]=2^{6k}(63)+3^{2k-2}(8)[/tex]
After this I'm stuck! I know that I have to write it in the form of [tex]5B[/tex](where B is a constant) but I cant. This is because if I do take 5 common I get fractions in the above expression.
Thanks in advance for any help.
[tex] f(n) \equiv 2^{6n}+3^{2n-2} [/tex] is divisible by 5. This is what I did:
Suppose that the given statement is true for [tex]n=k[/tex]
Since the[tex] f(k)[/tex] is divisible by 5,
[tex]f(k)=5A[/tex] (where A are is a constant.)
Also, from the given statement:
[tex] f(k)=2^{6k}+3^{2k-2} [/tex]
To prove that the given statement is also true for n=k+1:
[tex] f(k+1)-f(k) [/tex]
[tex]=2^{6k+6}+3^{2k} - (2^{6k}+3^{2k-2})[/tex]
[tex]=2^{6k}(63)+3^{2k-2}(8)[/tex]
After this I'm stuck! I know that I have to write it in the form of [tex]5B[/tex](where B is a constant) but I cant. This is because if I do take 5 common I get fractions in the above expression.
Thanks in advance for any help.
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