We want to show that the nth + 1 term of a sequence that is defined as [itex]x_{n+1} = b + ax_n, n\geq1 \ a,b \in \mathbb{R} [/itex] is given by(adsbygoogle = window.adsbygoogle || []).push({});

[tex] x_{n+1} = a^nx_1 + b\frac{1-a^n}{1-a} \ \mbox{if} \ a\neq 1 [/tex]

The manual suggests that we proceed by induction. Let us proceed by induction.

P(1) is that

[tex]x_{1+1}=x_2=a^1x_1+b\frac{1-a^1}{1-a}[/tex]

[tex]x_2=b+ax_1[/tex],

which is in agreement with the expression of [itex]x_2[/itex] implied by the definition of [itex]x_{n+1}[/itex]

Let us suppose P(n) to be true, i.e. that [itex]x_{n+1}[/itex] is in fact

[tex]x_{n+1}=a^nx_1+b\frac{1-a^n}{1-a}[/tex]

And let's see if P(n)being true implies that P(n+1) is true. P(n+1) is that

[tex]x_{n+1+1}=x_{n+2}=a^{n+1}x_1+b\frac{1-a^{n+1}}{1-a}[/tex]

but after "simplifing" to

[tex]x_{n+2}=aa^nx_1+b\frac{1-aa^n}{1-a}[/itex],

I don't see how to go any further than that!

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

Join Physics Forums Today!

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

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

# Proof by induction - I'm stuck!

Loading...

Similar Threads - Proof induction stuck | Date |
---|---|

Proof by mathematical induction | Apr 5, 2013 |

Proof by induction, puzzles by answer | Jun 6, 2012 |

Nth Derivative Induction Proof | Sep 26, 2009 |

Help proof by induction | Jan 21, 2009 |

Proof by Induction | Sep 17, 2008 |

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