F n - f n -1- if n is odd -2 f n -1- if n is even-If f 0-1- then find the value of f 20- f 21A- 2048B- 512C- 4096D- 128E- 1

The correct option is A 2048Soln:Option (b)f4 + f5 = 2f3 + f4 = 2f2 + 2f3 = 4f1 + 2f2 = 4f0 + 4f1 = 4f0 + 4f0 = 8f0 = 8Hence option (e)OR starting with f(0) = ...

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Quantitative Aptitude

Functions

f n = f n -1,...

Question

f_n= f_{n – 1}, if n is odd= 2f_n-1, if n is even.If f₀ = 1, then find the value of f₂₀ + f₂₁.

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2048

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512

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4096

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128

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Solution

The correct option is B 2048
Soln:
Option (b)
f₄ + f₅ = 2f₃ + f₄ = 2f₂ + 2f₃ = 4f₁ + 2f₂ = 4f₀ + 4f₁ = 4f₀ + 4f₀ = 8f₀ = 8
Hence option (e)
OR starting with
f(0) = 1
F(1) = 1
F(2) = 2
F( 3) = 2
F(4) = 4
F (5) = 4
F(6) = 8
F(7) = 8
……………..
…………….
F(N) =2^N/2
F(N+1) = 2^N/2
Therefore f( 20) = 2¹⁰and f(21) = 2¹⁰
f(20) + f(21) =1024+1024 = 2048

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fn= fn – 1, if n is odd= 2fn-1, if n is even.If f0 = 1, then find the value of f20 + f21.

f_n= f_{n – 1}, if n is odd= 2f_n-1, if n is even.If f₀ = 1, then find the value of f₂₀ + f₂₁.