Question

for n>= 1 use mathematical induction

Answer 1

\[15/2^{4n} -1\]

Answer 2

i got most of it im just stuck..

Answer 3

this is what i have

Answer 4

can you state the complete question...

Answer 5

yes give me a second

Answer 6

\(15 | (2^{4n} -1) \)

Answer 7

\[15|2^{4n}-1=15p\]
\[P(1): 15|2^{4(1)}-1\]
==>
\[15|2^4-1\]
=>\[15|15 true \]
P(k)\[15|2^{4k}-1<=>2^{4k}-1=15p\]
P(k+1)\[15|2^{4k+4}-1<=>2^{4k+4}-1=15m\]
\[2^4*2^k-1=15m\]

Answer 8

so then what it becomes?
\[16(2^{4k}-1=15m\]

Answer 9

Looks good !

Answer 10

ok but im stuck right here...

Answer 11

you multiplied 2^4 both sides, eh ?

Answer 12

now we need to prove on that 15 divides 2^(4*(k+1) - 1
using 2^4k -1 =15p

Answer 13

ok...

Answer 14

P(k) : \(2^{4k}-1=15p \)

multiply \(2^4\) both sides :

\(2^4\left(2^{4k}-1\right)=15q \)

\( 2^{4k+4}-16=15q \)

\(2^{4(k+1)}-1-15=15q \)

Answer 15

2^(4*(k+1) - 1 =2^4 *2^4k - 1
=2^4(15p+1) -1

Answer 16

\[2^{4(k+1)}-1=2^{4k}2^4-1=2^{4k}(2^4-1+1)-1\]

\[=2^{4k}(2^4-1)+2^{4k}-1\]

Answer 17

=15*16p +16 -1 = 15*16p +15=15(16p+1)

Answer 18

p is integer hence (16p+1) is also an integer

Answer 19

hence further...

Answer 20

2^[4*(k+1)] = 2^(4k+4)
= 2^4 * 2^(4k)
So now we have to show that

2^4 * 2^(4k) - 1 = 15m

2^4 * 2^(4k) - 1 =
2^4 * 2^(4k) - 16 + 15
= 16 * [2^(4k) - 1] + 15
= 16 * 15p + 15
= 16 * 15p + 15
= 15*m

Answer 21

I've showed that it lead to a multiple of 15
Start with
2^4 * 2^(4k) - 1
next rewrite the -1 as -16 + 15
then 2^4 * 2^(4k) - 16 + 15
2^4 = 16

So we get
16 * 2^(4k) - 16 + 15
Next we factor 16 from the first two terms
16 [2^(4k) - 1] + 15

Now the thing in square brackets is our assumption
[2^(4k) - 1] = 15p
So substituting that we get
16*15p + 15
Next we factor out the 15
15*[16p - 1]

So since 16p-1 is also an integer, I have shown through algebra that 2^[4(k+1)] - 1 = 15*[16p -1] = 15 * m
therefore 15 divides 2^[4(k+1)] - 1
where m is the integer equal to 16p-1

Answer 22

thanks for the detail thats what i needed! lol

Answer 23

ha ha, no worries. :-]