Proving 11 is a factor of 2^(4n+3)+3(5^n)

[charmedbeauty]

Homework Statement

Theorem if n is a non-negative integer then 11 is a factor of 2^(4n+3)+3(5^n)

Basic ideas if 2^(4n+3) =11k -3(5^n) then,
2^(4n+7) =11(16k -3(5^n))-3(5^(n+1))

Homework Equations

The Attempt at a Solution

So I understand how they get the result 2^(4n+3) =11k -3(5^n)

but I have no idea how to obtain the second result, namely, 2^(4n+7) =11(16k -3(5^n))-3(5^(n+1)) ?

Im stumped I can obviously see how they have added another 2^4 but I really don't see how that produces RHS. Help!

 

[Karamata]

Mathematical induction, $k \rightarrow k+1$

 

[charmedbeauty]

Mathematical induction, $k \rightarrow k+1$

Ok but when I do my basis step for n=1...

2^(4+3) +15 = 11k

so 2^7 +15 =11k

so 143= 11(13) which is true

so if 2^(4n+3)=11k -3(5^n)...

so 11k = 2^(4n+3) +3(5^n)

now check its true for n+1

so 2^(4(n+1)+3) =11K-3(5^(n+1))

so 2^(4n+7) = 11k-3(5^(n+1))

so subing in for 11k from above I have

2^(4n+7) = (2^(4n+3)+3(5^n))-3(5^(n+1))'

but I need to show

2^(4n+7)=11(16k-3(5^n))-3(5(n+1))

?

I have not really done any mathematical induction before I just watched a few videos, so I have probably done something wrong. Help!

so 1

 

[Curious3141]

This shouldn't be under Calculus.

Here's a hint:

$$2^{4(n+1) + 3} + 3.5^{n+1} = 16.2^{4n + 3} + 5.(3.5^n)$$

and 16 = 5 + 11. Regroup those factors! Try and manipulate it into something like:

$$p.(2^{4n + 3} + 3.5^n) + 11q$$

where p and q are integers or integral expressions that I'll leave you to determine.

Once you've assumed the inductive hypothesis that $(2^{4n + 3} + 3.5^n)$ is a multiple of 11, wouldn't that whole expression magically become a multiple of 11 too?

Then all you need to do is test it for n=1.