Math question (Algebra)

Question

If f(x)=2^(3x+1)

Show that f(x+1)-f(x) is divisible by 7 if x is a whole, non-negative number.

I tried solving this but I always block at the same place and have no clue how to proceed.

Answer 1

f(x+1) = 2^{3(x+1) + 1} = 2^{3x +4} = z

 

f(x) = z/8 because f(x+1) is always 2³ = 8 times less

 

f(x+1) - f(x) = 2^3x+4 - 2^3x+1 = z - z/8 = (7/8)z

 

divide (7/8)z by 7 to get z/8 = f(x) which is a whole number so the difference is divisible by 7

 

 

 

 

 

 

 

 

 

 

Answer 2

f(x+1) = 2^(3x +4)

f(x+1) - f(x) is 2^(3x +4)  - 2^(3x +1)

 

Factor out a common factor of 2^(3x +1) and you'll get.

2^(3x +1) *[2^3 -1] or 2^(3x +1) *7

 So this expression is always a multiple of 7 and is therefore always divisible by 7