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.
Hello, rcs!
If you can read the problem, you can baby-talk your way through it.
$\displaystyle \text{Let }f(x)\text{ be a function de{f}ined on the set of integers}$
$\displaystyle \text{such that: }\:f(1) = 5\,\text{ and }\,f(x\!+\!1) \:=\: 2\!\cdot\!f(x) + 1\,\text{ for all integers }x.$
$\displaystyle \text{What is the value of: }\:f(7) - f(0)\,?$
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: .$\displaystyle f(7) - f(0) \;=\;383 -2 \;=\;381$