induction q (1 Viewer)

mmmmmmmmaaaaaaa · Oct 18, 2022

try subbing in that u=2^n +5^n into n=k+1

Longelbow89 · Oct 18, 2022

Deem_Skills said:
View attachment 36631

I’m pretty sure recursive induction is ext 2 only, and the syllabus says it’s only first order recursive formulae where that’s a second order recursive so I don’t think that’s necessary to know, even for ext 2

Drongoski · Oct 18, 2022

Can show to be true for n=3

$\text {Now assume true for } 3 \leq n \leq k , $ and $ k \geq 3 \text {(Strong Law)}\\ \\ \therefore U_k = 2^k + 5^k \\ \\ \therefore U_{k+1} = 7U_k - 10 U_{k-1} = 7(2^k + 5^k) - 10(2^{k-1} + 5^{k-1}) \\ \\ = 7\times2^k + 7\times 5^k - 10 \times 2^{k-1} - 10 \times 5^{k-1} \\ \\ =7\times2^k + 7\times 5^k -5\times 2^k - 2 \times 5^k \\ \\ = (7-5)2^k + (7-2)5^k \\ \\ = 2^{k+1} + 5^{k+1}$

Therefore . . . blah .. blah .. blah . . .

QED

induction q (1 Viewer)

Deem_Skills

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mmmmmmmmaaaaaaa

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Longelbow89

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Drongoski

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