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Question

fn=fn1, if n is odd = 2fn1, if n is even.If f0 = 1, then find the value of f20+f21.
  1. 128
  2. 2048
  3. 512
  4. 4096


Solution

The correct option is B 2048

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) =2N2

F(N+1) = 2N2

Therefore f( 20) = 210 and f(21) = 210

f(20) + f(21) =1024+1024 = 2048

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