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 21-A- 128B- 2048C- 512D- 4096

The correct option is B 2048f(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 ...

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Quantitative Aptitude

Functions

f n = f n -1,...

Question

$f_{n} = f_{n - 1}$ , if n is odd = $2 f_{n - 1}$ , if n is even.If $f_{0}$ = 1, then find the value of $f_{20} + f_{21} .$

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 A 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) = $2^{\frac{N}{2}}$

F(N+1) = $2^{\frac{N}{2}}$

Therefore f( 20) = $2^{10}$ and f(21) = $2^{10}$

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

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Similar questions

Q. If

f (1) = 1, f (n + 1) = 2 f (n) + 1, n \geq 1,

then

f (n)

is :

f (1) = 1, n \geq 1 \Rightarrow f (n + 1) = 2 f (n) + 1

then

f (n) =

Q. If

f (x)

is a polynomial of degree

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such that

f (0) = 0, f (1) = \frac{1}{2}, . . . ., f (n) = \frac{n}{n + 1}

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f (n + 1)

Q. If

f (n + 1) = f (n) + n

for all

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equals

If f( n + 1) = f (n) + n for all n ≥ 0 or f (0) = 1 then f (200) equals

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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.

The correct option is A 2048f(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

$f_{n} = f_{n - 1}$ , if n is odd = $2 f_{n - 1}$ , if n is even.If $f_{0}$ = 1, then find the value of $f_{20} + f_{21} .$

The correct option is A 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) = $2^{\frac{N}{2}}$

F(N+1) = $2^{\frac{N}{2}}$

Therefore f( 20) = $2^{10}$ and f(21) = $2^{10}$

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