Proof By Mathematical Induction

**ah-bee** · May 6th 2008, 01:44 AM

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.

**Soroban** · May 6th 2008, 05:29 AM

Hello, ah-bee!

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

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

(a) Use Induction to prove that the sequence is bounded above by 6.

Verify $S(1)\!:\;\;a_1\,=\,6\,\leq 6$ . . . True!

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

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

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

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

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

**ah-bee** · May 6th 2008, 05:32 AM

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

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