Respuesta :
A is obviously false.
E is false because it says "exactly 1.618" and the ratio between successive terms varies but approaches the golden number (1 + sqrt5)/2 which is irrational and not exactly 1.618
B is correct. By the way, showing that the first four are multiples of 3 is no proof.
Proof.
F(n + 4) = F(n + 3) + F(n + 2)
= [F(n + 2) + F(n + 1)] + F(n + 2)
= 2*F(n + 2) + F(n + 1)
= 2*[F(n + 1) + F(n)] + F(n + 1)
= 3*F(n + 1) + 2*F(n)
= 3*[F(n) + F(n - 1)] + 2*F(n)
= 5*F(n) + 3*F(n - 1)
If F(n) = 3m, i.e. a multiple of 3 then
F(n + 4) = 3m + 3*F(n - 1) = 3(m + F(n - 1), i.e. also a multiple of 3.
C is correct. In fact Sum [i = 1 to n] F(i) = F(n + 2) - 1 for all n. Proof by induction.
Clearly true for n = 2 because
Sum [i = 1 to 2] F(i) = F(1) + F(2) = 2 = F(4) - 1
Sum [i = 1 to n + 1] F(i) = {Sum [i = 1 to n] F(i)} + F(n + 1)
= F(n + 2) - 1 + F(n + 1) by the assumption true for n
= F(n + 2) + F(n + 1) - 1
= F(n + 3) - 1 by the Fibonacci formula
= F((n + 1) + 2) - 1
which is the same formula with n replaced by n + 1
Hence true for all n by mathematical induction.
D is correct. Proof by induction. I'll leave that to you.
E is false because it says "exactly 1.618" and the ratio between successive terms varies but approaches the golden number (1 + sqrt5)/2 which is irrational and not exactly 1.618
B is correct. By the way, showing that the first four are multiples of 3 is no proof.
Proof.
F(n + 4) = F(n + 3) + F(n + 2)
= [F(n + 2) + F(n + 1)] + F(n + 2)
= 2*F(n + 2) + F(n + 1)
= 2*[F(n + 1) + F(n)] + F(n + 1)
= 3*F(n + 1) + 2*F(n)
= 3*[F(n) + F(n - 1)] + 2*F(n)
= 5*F(n) + 3*F(n - 1)
If F(n) = 3m, i.e. a multiple of 3 then
F(n + 4) = 3m + 3*F(n - 1) = 3(m + F(n - 1), i.e. also a multiple of 3.
C is correct. In fact Sum [i = 1 to n] F(i) = F(n + 2) - 1 for all n. Proof by induction.
Clearly true for n = 2 because
Sum [i = 1 to 2] F(i) = F(1) + F(2) = 2 = F(4) - 1
Sum [i = 1 to n + 1] F(i) = {Sum [i = 1 to n] F(i)} + F(n + 1)
= F(n + 2) - 1 + F(n + 1) by the assumption true for n
= F(n + 2) + F(n + 1) - 1
= F(n + 3) - 1 by the Fibonacci formula
= F((n + 1) + 2) - 1
which is the same formula with n replaced by n + 1
Hence true for all n by mathematical induction.
D is correct. Proof by induction. I'll leave that to you.
Answer:
B E and D
Step-by-step explanation:
B.) The fourth Fibonacci number is 3. Every fourth Fibonacci number after that is divisible by 3.
D.) Starting with F11, the difference between any Fibonacci number and the Fibonacci number that comes 10 before it (for example, F37-F27) is a number evenly divisible by 11.
E.) The sum of the first n Fibonacci numbers is always exactly 1 less than another Fibonacci number.
