Let f be a function dened on the set of integers such that f(1) = 5

and f(x + 1) = 2f(x) + 1 for all integers x. What is the value of

f(7) - f(0)?

i bet the answer is 381 but i couldn't make how to the solution is done.

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- Nov 24th 2010, 05:54 PMrcsfunction equation functions
Let f be a function dened on the set of integers such that f(1) = 5

and f(x + 1) = 2f(x) + 1 for all integers x. What is the value of

f(7) - f(0)?

i bet the answer is 381 but i couldn't make how to the solution is done. - Nov 24th 2010, 05:59 PMmr fantastic
- Nov 24th 2010, 06:20 PMpickslides
- Nov 25th 2010, 05:10 AMrcs
a friend of mine told me that it is 381,,, but she couldnt show to me how she solved it,,, that is why i just guessed it. hope i could get the right flow of the solution. thanks

- Nov 25th 2010, 05:17 AMPlato
- Nov 25th 2010, 05:57 AMSoroban
Hello, rcs!

If you can read the problem, you can baby-talk your way through it.

Quote:

$\displaystyle \text{Let }f(x)\text{ be a function de{f}ined on the set of integers}$

$\displaystyle \text{such that: }\:f(1) = 5\,\text{ and }\,f(x\!+\!1) \:=\: 2\!\cdot\!f(x) + 1\,\text{ for all integers }x.$

$\displaystyle \text{What is the value of: }\:f(7) - f(0)\,?$

The first term is 5.

Each subsequent term is twice the preceding term, plus 1.

We have:

. . $\displaystyle \begin{array}{ccccc}

f(1) &=& 5 \\

f(2) &=&2(5)+1 &=& 11\\

f(3) &=& 2(11)+1 &=& 23 \\

f(4) &=& 2(23)+1 &=& 47 \\

f(5) &=& 2(47)+1 &=& 95 \\

f(6) &=& 2(95)+1 &=& 191 \\

f(7) &=& 2(192)+1 &=& 383 \end{array}$

What is $\displaystyle f(0)$ ?

Since $\displaystyle f(1) \:=\:2\!\cdot\!f(0) + 1 \:=\:5$, we have: .$\displaystyle f(0) = 2$

Therefore: .$\displaystyle f(7) - f(0) \;=\;383 -2 \;=\;381$

- Nov 25th 2010, 10:47 PMrcs
thanks you so so much Soroban!!! ilove you all here! :) i can now have a good night sleep