1. ## Induction

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.

2. Originally Posted by joksy
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.
prove that 3 < a,, < 4 ??

what is a,, ?

3. Hello, joksy!

Let $a_0 = 3$, and let $a_{n+1} \:=\:\sqrt{a_n + 7}$

Prove that: . $3 < a_n < 4,\:\text{ for all }n > 0$

Verify $S(1)\!:\;\;a_1 \:=\:\sqrt{a_0+7} \:=\:\sqrt{3 + 7} \:\approx\:3.16$ . . . True

Assume $S(k)\!:\;\;3 \:<\:a_k \:<\:4$

Add 7: . $10 \:<\:a_k + 7 \:<\:11$

$\text{Take square roots: }\;\sqrt{10} \:<\:\underbrace{\sqrt{a_k+7}}_{\text{This is }a_{k+1}} \:<\:\sqrt{11}$

So we have: . $3.162 \:<\:a_{k+1} \:<\:3.317 \quad\hdots\quad 3 \:<\:a_{k+1} \:<\:4$

We have shown that $S(k+1)$ is true.
. . The inductive proof is complete.

4. doh. Thanks a lot jhevon, I did see this trick... I have to work a lot on more exercise !! But anyway thanks a lot

5. Originally Posted by joksy
doh. Thanks a lot jhevon, I did see this trick... I have to work a lot on more exercise !! But anyway thanks a lot
well, it was Soroban that helped you

6. Sorry where a red your name... so THANKS SOROBAN and sorry for the misunderstanding. and for you thanks for tell me the error.

Have a nice day guys