1. Mathematical Induction

Plz help on these:

(a) Prove by mathematical induction that
$6^{n+2} + 7^{2_{n+1}}$ is divisible by 43 for each positive integer n

(b) Show that $n^2 > 2n +1$ for $n \ge 3$ using mathematical induction method.

2. Originally Posted by cu4mail
Plz help on these:

(a) Prove by mathematical induction that
$6^{n+2} + 7^{2_{n+1}}$ is divisible by 43 for each positive integer n
i'll start you off...

we wish to prove the statement:

$P(n):6^{n + 2} + 7^{2n + 1}$ is divisible by 43 for all $n \in \mathbb{N}$.

$P(1)$ is true, so we have our inductive step.

Assume then that $P(k)$ is true for $k \ge 1, k \in \mathbb{N}$, we show that $P(k + 1)$ is true.

Now, $P(k + 1) = 6^{k + 3} + 7^{2k + 3} = 6^{k + 3} + \underbrace{6 \cdot 7^{2k + 1} - 6 \cdot 7^{2k + 1}}_{\mbox{This is zero}} + 7^{2k + 3}$
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.
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now continue

3. Originally Posted by cu4mail
(b) Show that $n^2 > 2n +1$ for $n \ge 3$ using mathematical induction method.
this one is not that bad, in fact, i think giving you a hint would give the answer away anyway, so let's just do it.

we wish to prove the statement:

$P(n): n^2 > 2n + 1$ for all $n \ge 3, n \in \mathbb{N}$

Now, $P(3): 9 > 7$ which is true. So the inductive step is complete.

Assume $P(k)$ is true for some $k \ge 3, k \in \mathbb{N}$, we show $P(k + 1)$ is true. that is, we want to show $(k + 1)^2 > 2(k + 1) + 1$

now, $P(k + 1): (k + 1)^2 = k^2 + 2k + 1$

$> 2k + 1 + 2k + 1$ .....since $k^2 > 2k + 1$

$\ge 2k + 1 + 6 + 1$ ......since $k \ge 3$

$= 2k + 8$

$> 2k + 3$

$= 2(k + 1) + 1$

thus we have shown $P(k + 1)$ is true.

Thus we have proven $P(n)$ by induction