Powers of 2 Go In, Powers of 2 Come Out

Algebra Level 3

Let $f$ be a function with the following properties:

$f(1) = 1$
$f(2n) = n \cdot f(n)$ for all positive integers $n$

What is the value of $f(2^{100})$ ?

2^{4950}

2^{9999}

2^{999}

2^{100}

2 solutions

Rishabh Jain
Jan 16, 2016

$f(2^{100})=f(2\times \color{#D61F06}{2^{99}})$ $=\color{#D61F06}{2^{99}}\times f(2\times \color{#D61F06}{2^{98}})$ $=\color{#D61F06}{2^{99+98}}\times f(2\times \color{#D61F06}{2^{97}})$ $.$ $.$

$=\color{#D61F06}{2^{99+98+97....+2+1+0}}\times f(\color{#D61F06}{2^{0}})$ $\Large =2^\frac{99\times 100}{2}=\color{limegreen}{ 2^{4950}}$

Hey how do u use colours in ur solutions?

Atul Shivam - 5 years, 5 months ago

Did exactly the same way

Atul Shivam - 5 years, 5 months ago

Ved Sharda
Jan 16, 2016

Given : f (2n) = n * f(n)

f(2^100) = [2^99] * f(2^99)

Now again f(2^99) = [2^98] * f(2^98)

similarly this process continues and we get

f(2^100) = [2^99] * [2^98] * [2^97] ............ [2^2] * [2^1] * 1

f(2^100) = [2^(99+98+97+......+2+1)]

we know that 99+98+97......+2+1 = 99(99+1)/2

so 99+98+97+......+2+1 = 4950

so f(2^100) = 2^4950

IS THIS METHOD CORRECT ?

Powers of 2 Go In, Powers of 2 Come Out

2 solutions

0 pending reports