Hi Guys I have a question for you.
The statement is :
Let ao = 3, and let a_(n+1) = (an + 7)^(1/2) if n > 0. Prove that 3 < a„ < 4
for all n > 0.
Thanks a lot for the help.
Hello, joksy!
Let $\displaystyle a_0 = 3$, and let $\displaystyle a_{n+1} \:=\:\sqrt{a_n + 7}$
Prove that: .$\displaystyle 3 < a_n < 4,\:\text{ for all }n > 0$
Verify $\displaystyle S(1)\!:\;\;a_1 \:=\:\sqrt{a_0+7} \:=\:\sqrt{3 + 7} \:\approx\:3.16$ . . . True
Assume $\displaystyle S(k)\!:\;\;3 \:<\:a_k \:<\:4$
Add 7: . $\displaystyle 10 \:<\:a_k + 7 \:<\:11 $
$\displaystyle \text{Take square roots: }\;\sqrt{10} \:<\:\underbrace{\sqrt{a_k+7}}_{\text{This is }a_{k+1}} \:<\:\sqrt{11}$
So we have: .$\displaystyle 3.162 \:<\:a_{k+1} \:<\:3.317 \quad\hdots\quad 3 \:<\:a_{k+1} \:<\:4$
We have shown that $\displaystyle S(k+1)$ is true.
. . The inductive proof is complete.