# Thread: Proof By Mathematical Induction

1. ## Proof By Mathematical Induction

The sequence (an) is defined inductively (or recursively) by an1 = 6 and an+1 = (sqrt)(4an + 5) for n >= 1.
(a) Use the Principle of Mathematical Induction to prove that the sequence (an) bounded above by 6.
(b) Use the Principle of Mathematical Induction to prove that the sequence (an) is bounded below by 5.
(c) Use the Principle of Mathematical Induction to prove that the sequence (an) is decreasing.

I have done it already.. but it kinda seems too simple, so I wanna check with the solutions of other people.

2. Hello, ah-bee!

I think you're right . . . It is rather simple.

The sequence $\displaystyle a_n$ is defined recursively by: .$\displaystyle a_1 = 6\,\text{ and }\,a_{n+1} = \sqrt{4a_n + 5}\,\text{ for }n \geq 1$

(a) Use Induction to prove that the sequence is bounded above by 6.
Verify $\displaystyle S(1)\!:\;\;a_1\,=\,6\,\leq 6$ . . . True!

Assume $\displaystyle S(k)\!:\;\;a_k\:\leq\:6$

. . Multiply by 4: .$\displaystyle 4a_k \:\leq \:24$

. . Add 5:. . $\displaystyle 4a_k + 5 \:\leq \:29$

. . $\displaystyle \text{Take the square root: }\;\underbrace{\sqrt{4a_k + 5}}_{\text{This is }a_{k+1}} \:\leq \:\sqrt{29} \:<\:6$

We have shown that: .$\displaystyle a_{k+1}\:\leq\:6$
The inductive proof is compete.

3. yeah thats what i got... seems a bit too simple. thanks for verification