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junior college 1 | H2 Maths
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Hello! Please help me with this question :)
Date Posted: 3 years ago
Views: 332
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Q5a
Date Posted:
3 years ago
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Second one, but not so sure about this one
Date Posted:
3 years ago
Fun fact :
The common ratio (√5 - 1)/2 is actually the reciprocal of the golden ratio φ ,(1 + √5)/2
The common ratio (√5 - 1)/2 is actually the reciprocal of the golden ratio φ ,(1 + √5)/2
Actually, I was thinking that the final value of the expression I answered was the golden ratio itself. Seems like I got the values mixed up.
The conjugate, (-1 - √5)/2 is the negative of the golden ratio
Another fun fact :
(y/x)² + (y/x) - 1 = 0
(y/x)² = 1 - (y/x) ①
The 1st term of the G.P = x²
2nd term = x² (y/x)
3rd term = x² (y/x)²
= x² (1 - y/x)
= x² - x² (y/x) (from ①)
= 1st term - 2nd term
4th term
= x² (y/x)³
= x² (y/x)² (y/x)
= (x² - x² (y/x)) (y/x)
= x²(y/x) - x²(y/x)²
= 2nd term - 3rd term
Likewise,
5th term = 3rd term - 4th term
6th term = 4th term - 5th term
...
...
nth term = (n-2)th term - (n-1)th term
This can be rewritten as :
nth term + (n-1)th term = (n-2)th term
This G.P is actually a generalised Fibonacci sequence, but in reverse order.
(y/x)² + (y/x) - 1 = 0
(y/x)² = 1 - (y/x) ①
The 1st term of the G.P = x²
2nd term = x² (y/x)
3rd term = x² (y/x)²
= x² (1 - y/x)
= x² - x² (y/x) (from ①)
= 1st term - 2nd term
4th term
= x² (y/x)³
= x² (y/x)² (y/x)
= (x² - x² (y/x)) (y/x)
= x²(y/x) - x²(y/x)²
= 2nd term - 3rd term
Likewise,
5th term = 3rd term - 4th term
6th term = 4th term - 5th term
...
...
nth term = (n-2)th term - (n-1)th term
This can be rewritten as :
nth term + (n-1)th term = (n-2)th term
This G.P is actually a generalised Fibonacci sequence, but in reverse order.
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