1. ## Induction divisibility P2

Could somebody help me with these questions? I just need to prove that n=k+1 using the assumption for n=k

Prove:
1) 7^2n - 48n -1 for 2304 for n=1,2,3...
2) (3n+1).7^n -1 is divisible by 9 for n=1,2,3...
3) 10^n +3[4^(n+2)] +5 is divisible by 9 for n=1,2,3...

Thank you !

2. $
(3n + 1)\cdot 7^n - 1 = 9k; k \in \mathbb{N}
$

basis:
$
(3(1) + 1)\cdot 7^{(1)} - 1 = 9k
$

$
27 = 9k
$

$
k = 3
$

hypothesis:
$
(3n + 1)\cdot 7^n - 1 = 9k; k \in \mathbb{N}
$

$
(3(n + 1) + 1)\cdot 7^{n + 1} - 1 = 9k; k \in \mathbb{N}
$

$
7\cdot 7^n(3n + 1 + 3) - 1 = 9k; k \in \mathbb{N}
$

$
7\cdot 7^n(3n + 1) - 1 + 7\cdot 7^n\cdot 3 = 9k; k \in \mathbb{N}
$

finished. We have 7 times a multiple of 9 and 7 times a multiple of 63, which is also divisble by 9.
* actually there is an error here- can someone do n = 2 for me?

for 3)
$
9\cdot 10^n + 10^n + 9\cdot 4^{m + 2} + 3\cdot4^{n + 2} + 5 = 9k
$

3. Originally Posted by christina
Could somebody help me with these questions? I just need to prove that n=k+1 using the assumption for n=k

Prove:
1) 7^2n - 48n -1 for 2304 for n=1,2,3... can you resubmit this?
2) (3n+1).7^n -1 is divisible by 9 for n=1,2,3...
3) 10^n +3[4^(n+2)] +5 is divisible by 9 for n=1,2,3...

Thank you !
2)

$\frac{(3n+1)7^n-1}{9}=natural\ number$

$\frac{[3(n+1)+1]7^{n+1}-1}{9}=\frac{\left([3n+1+3]7^n\right)7-1}{9}$

$=\frac{\left((3n+1)7^n\right)7+3(7)7^n-1}{9}$

$=\frac{(3n+1)7^n-1+(3n+1)7^n(6)+7(3)7^n}{9}$

$=\frac{(3n+1)7^n-1+(18n)7^n+(6)7^n+(21)7^n}{9}$

$=\frac{(3n+1)7^n-1+(18n)7^n+(27)7^n}{9}$

As the other 2 terms of the numerator are divisible by 9,
then if the first term is divisible by 9, the hypothesis is proven if it's true for n=1.

3.

$\frac{10^n+3\left(4^{n+2}\right)+5}{9}=natural\ number$

$\frac{10^{n+1}+3\left(4^{n+3}\right)+5}{9}=\frac{( 10)10^n+3\left(4^{n+2}\right)+(9)10^n+(3)3\left(4^ {n+2}\right)+5}{9}$

$=\frac{\left(10^n+3\left(4^{n+2}\right)+5\right)+9 \left(10^n+4^{n+2}\right)}{9}$

Therefore, if

$10^n+3\left(4^{n+2}\right)+5$

is divisible by 9

$10^{n+1}+3\left(4^{n+3}\right)+5$

is also divisible by 9.

Hence check foe n=1.