#1**-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...

ProffesorNobody Jun 12, 2019

#4**+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.

Guest Jun 12, 2019

edited by
Guest
Jun 12, 2019

#5**+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

CPhill Jun 12, 2019