Question

the sequence of f1=1, f2=3, fn+2=fn+fn+1 for n>1.how many even terms are b/w the first 2020 terms of that sequence

Answer 1

SOLUTION

TO DETERMINE

The number of even terms between the first 2020 terms of the sequence

$\sf{f_1= 1, f_2 = 3, f_{n+2} = f_n + f_{n+1}}$

EVALUATION

Here the given sequence is

$\sf{f_1= 1, f_2 = 3, f_{n+2} = f_n + f_{n+1}}$

Thus we have

$\sf{f_1= 1}$

$\sf{ f_2 = 3}$

$\sf{ f_{3} = f_1 + f_{2} = 1 + 3 = 4}$

$\sf{ f_{4} = f_2 + f_{3} = 3 + 4 = 7}$

$\sf{ f_{5} = f_3 + f_{4} = 4+ 7 = 11}$

$\sf{ f_{6} = f_4 + f_{5} = 7+ 11 = 18}$

$\sf{ f_{7} = f_5+ f_{6} = 11+ 18 = 29}$

So on

Thus we see that in every three terms there is exactly one even number

Hence the required number of even terms between the first 2020 terms of the sequence

$\displaystyle \sf{ = \bigg[ \frac{2020}{3} \bigg]}$

= [ 673.33 ]

= 673

Where [ x ] denotes the greatest integer but not greater than x

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Answer 2

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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