Induction Help (1 Viewer)

Francis006 · Feb 17, 2024

The equation x^2+x+1=0 has roots a and b. Also, a series is defined by Tn=a^n+b^n for n=1,2,3,4...

a) Show that T1=-1 and T2= -1 (completed, but included for context)
b) Show that Tn = -Tn-1 - Tn-2 for n=3,4,5... (I have no clue from here)
c) Hence use induction to show that Tn=2cos(2nπ/3) for n=1,2,3...
d) Hence write down the value of Σn=2012 k=1 Tk

Luukas.2 · Feb 17, 2024

Part (b) is a LHs = ... and RHS = ...

$\begin{align*} \text{By definition:} \qquad \text{LHS} &= T_n \\ &= a^n + b^n \qquad \text{where $a$ and $b$ are the roots of $x^2 + x + 1 = 0$} \\ \\ \text{RHS} &= -T_{n-1} - T_{n-2} \qquad \text{provided $n > 2$ and $n \in \mathbb{Z}$} \\ &= -\left(a^{n-1} + b^{n-1}\right) - \left(a^{n-2} + b^{n-2}\right) \qquad \text{by definition} \\ &= -a^{n-2}\left(a + 1\right) - b^{n-2}\left(b + 1\right) \\ &= -a^{n-2} \times -a^2 - b^{n-2} \times -b^2 \qquad \text{as both $x = a$ and $x = b$ satisfy the equation $x + 1 = -x^2$} \\ &= a^n + b^n \\ &= \text{LHS} \qquad \text{as required} \end{align*}$

Part (c) will require strong induction:

Prove the result for $n = 1$ and $n = 2$
Assume it is true for $n = k$ and $n = k + 1$
Use that assumption with the definition of the recurrence, $T_{k+2} = -T_{k+1} - T_{k}$ to establish the result for $n = k + 2$ .

Part (d), the sum, can likely be done with a telescoping series-type approach or by establishing the repeating pattern of values of the sequence - the latter approach will be easier in this case (in the sense of being more obvious), though the former is quicker.

Francis006 · Feb 17, 2024

Thank You

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Induction Help (1 Viewer)

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Luukas.2

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