Heres the paper but i am stuck on question 3B
#Fibonacci sequence
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astral flowerBOT
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first, lets find out what F(n+2) , F(n+1) , F(n) are:
$F(n+2) = \frac{\alpha^{n+2} - \beta^{n+2}}{\sqrt{5}}\$
$F(n+1) = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}}\$
$F(n) = \frac{\alpha^{n+2} - \beta^{n}}{\sqrt{5}}$
peak kiteBOT
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now we are going to find the sum of F(n+1) + F(n):
$F(n+1) + F(n) = \frac{\alpha^{n+1} + \alpha^{n} - \beta^{n+1} - \beta^{n}}{\sqrt{5}}$
peak kiteBOT
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here, we can use exponent rules to simplify the numerator:
$= \frac{\alpha^{n} \cdot \alpha + \alpha^{n} \cdot 1 - \beta^{n} \cdot \beta - \beta^{n} \cdot 1}{\sqrt{5}}\ = \frac{\alpha^{n} \cdot (\alpha + 1) - \beta^{n} \cdot (\beta + 1)}{\sqrt{5}}$
peak kiteBOT
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as you have already proven in question 1 , you can do this:
$x^{2} = x + 1 \quad \to \quad \alpha^{2} = \alpha +1 \quad , \quad \beta^{2} = \beta+1$
peak kiteBOT
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and now the last step is to plug in your values:
$F(n+1) + F(n) = \frac{\alpha^{n} \cdot (\alpha^{2}) - \beta^{n} \cdot (\beta^{2})}{\sqrt{5}}\
= \frac{\alpha^{n+2} - \beta^{n+2}}{\sqrt{5}} = F(n+2)$
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$F(n+2) = \frac{\alpha^{n+2} - \beta^{n+2}}{\sqrt{5}}\$
$F(n+1) = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}}\$
$F(n) = \frac{\alpha^{n} - \beta^{n}}{\sqrt{5}}\$
$F(n+1) + F(n) = \frac{\alpha^{n+1} + \alpha^{n} - \beta^{n+1} - \beta^{n}}{\sqrt{5}}\$
$= \frac{\alpha^{n} \cdot \alpha + \alpha^{n} \cdot 1 - \beta^{n} \cdot \beta - \beta^{n} \cdot 1}{\sqrt{5}}\$
$= \frac{\alpha^{n} \cdot (\alpha + 1) - \beta^{n} \cdot (\beta + 1)}{\sqrt{5}}\$
$x^{2} = x + 1 \quad \to \quad \alpha^{2} = \alpha +1 \quad , \quad \beta^{2} = \beta+1\$
$F(n+1) + F(n) = \frac{\alpha^{n} \cdot (\alpha^{2}) - \beta^{n} \cdot (\beta^{2})}{\sqrt{5}}\$
$= \frac{\alpha^{n+2} - \beta^{n+2}}{\sqrt{5}} = F(n+2)$
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this is the full length of the calculation without my notes