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 Prove that 5 n  + 6*7 n  +1 is divisible by 8 for all nonnegative integers.

 Jun 12, 2019
 #1
avatar+26 
-5

Ok, this isn't an answer... just some guessing, and randomness...

 

5n+42n+1

47n+1

ok so that thing simplified is 47n+1

then my brilliant idea is to get stuck. Somebody help....

 

EDIT: k, plug in numbers..... start with one

47+1=48

48 is divisible by 8.

 

next one: 95

um 95 isnt divisible by 8.... uhh what? did i do wrong...

 Jun 12, 2019
edited by ProffesorNobody  Jun 12, 2019
 #2
avatar+26 
-5

im sorry... im not really helping

ProffesorNobody  Jun 12, 2019
 #3
avatar+101390 
+2

You didn't do anything wrong.....this isn't true for all non-negative integers

 

 

cool cool cool

CPhill  Jun 12, 2019
 #4
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+2

I think that this is supposed to be    \(\displaystyle 5^{n}+6.7^{n}+1,\)

so let

\(\displaystyle f(n) = 5^{n}+6.7^{n}+1,\)

then

\(\displaystyle f(n+1)=5^{n+1}+6.7^{n+1}+1=5.5^{n}+42.7^{n}+1,\)

in which case

\(\displaystyle f(n+1)-5f(n)=(5.5^{n}+42.7^{n}+1)-5(5^{n}+6.7^{n}+1) \\ \hspace{72pt}=12.7^{n}-4,\)

so

\(\displaystyle f(n+1)=5f(n)+4(3.7^{n}-1).\)

 

The second term on the rhs is a multiple of 8, 

(3.7^n is odd so 3.7^n - 1 is even, and that's multiplied by 4)

so if the first term on the rhs is also a multiple of 8 the term on the lhs will be a multiple of 8.

 

i.e. if f(n) is divisible by 8 then f(n + 1) will also be divisible by 8.

 

Since f(1) = 5 + 6.7 + 1 = 48 is divisible by 8 it follows that f(2), (= 5^2 + 6.7^2 + 1 = 25 + 6.49 + 1 = 320) will be divisible by 8 and so then will

f(3), (= 5^3 + 6.7^3 + 1 = 125 + 6.343 + 1 = 125 + 2058 + 1 = 2184), etc.. 

 

f(n) will be divisible by 8 for all positive integers.

 Jun 12, 2019
edited by Guest  Jun 12, 2019
 #5
avatar+101390 
+4

5^n + 6*7^n+ 1

Show that it's true for  n = 1

5^1 + 6*7^1 + 1  =

5 + 42 + 1  =  48     which is divisible by 8

 

Assume that it's true for  k ...that is

5^(k) + 6*7^k + 1     is divisible by 8

 

Prove that it is true for k + 1

 

5^(k + 1)  + 6*7^(k + 1) + 1

 

5*5^k + 7*6*7^k + 1

 

5*5^k + 42* 7^k + 1

 

5^k + 4*5^k + 6*7^k + 36*7^k + 1

 

(5^k + 6*7^k + 1)  +   4*5^k + 36*7^k

 

(5^k + 6*7^k + 1) +  4*(5^k + 9*7^k)

 

Note that  5^k  is odd   and so is 9*7^k  ....but the sum of two odds is even = 2n

 

So we have

 

(5^k + 6*7^k + 1) + 4 (2n)  =

 

(5^k + 6*7^k + 1)  + 8n

 

The first was assumed to be divisible by 8   and the second term is also divisible by 8

 

 

cool cool cool

 Jun 12, 2019
edited by CPhill  Jun 12, 2019
 #6
avatar+26 
-3

ahhhhhhhhhhhhhhhhhhhhhhhh

ProffesorNobody  Jun 13, 2019

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