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!

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# Proof by induction - I'm stuck!

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