The sequence fn is given as f1=1,f2=3 , fn+2=fn+fn+1 for n_>1. How many even terms are between the first 2020
terms of that sequence?
Answers
Given : The sequence is given as , f1=1,f2=3,fn+2=fn+fn+1, for n>=1
To Find : How many even terms are between the first 2020 terms of that sequence
Solution:
f₁ = 1
f₂ = 3
f₃ = f₁ + f₂ = 1 + 3 = 4
f₄ = f₂ + f₃ = 3 + 4 = 7
f₅ = f₃ + f₄ = 4 + 7 = 11
f₆ = f₄ + f₅ = 11 + 7 = 18
f₇ = 29
f₈ = 47
f₉ = 76
Notice that for 3 terms there is one even term
2020 = 3 * 673 + 1
Hence there will 673 even terms
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Answer:
Step-by-step explanation:
Given : The sequence is given as , f1=1,f2=3,fn+2=fn+fn+1, for n>=1
To Find : How many even terms are between the first 2020 terms of that sequence
Solution:
f₁ = 1
f₂ = 3
f₃ = f₁ + f₂ = 1 + 3 = 4
f₄ = f₂ + f₃ = 3 + 4 = 7
f₅ = f₃ + f₄ = 4 + 7 = 11
f₆ = f₄ + f₅ = 11 + 7 = 18
f₇ = 29
f₈ = 47
f₉ = 76
Notice that for 3 terms there is one even term
2020 = 3 * 673 + 1
Hence there will 673 even terms