The question is: Prove by mathematical Induction that(adsbygoogle = window.adsbygoogle || []).push({});

[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.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proof by mathematical Induction: Divisibility

Loading...

Similar Threads - Proof mathematical Induction | Date |
---|---|

Proof Question: Using Mathematical Induction | Mar 24, 2008 |

Proof Question: Mathematical Induction | Mar 22, 2008 |

Proof by mathematical induction | Feb 17, 2008 |

Mathematical Induction and proofs | Feb 6, 2007 |

Proof by strong mathematical induction, can you see if this is enough to conclude? | Oct 2, 2006 |

**Physics Forums - The Fusion of Science and Community**