maths help

https://imgur.com/a/1gtnccI
im very confused on why they took away 3f(k) on this quesiton instead of 1, i get that when you take away 3 you can factorise it to be divisible by 5, but what indication is there that you need to take away 3f k? and surely taking away a different amount of f(k)'s gives different answers on its divisibility? I did the question taking away 1 f(k) and got different numbers obviously, but is it still divisible by 5? and if so how can i prove it since i cant factorise it? the solution bank doesnt seem to explain why they are taking away 3f(k). thank you

Reply 1

username1732133

Original post by Gent2324

I wouldn't worry about the way they did it.

You could write

8^{k+1}-3^{k+1}=8 \times 8^k - 3 \times 3^k = 5 \times 8^k + 3 \times 8^k - 3 \times 3^k = 5 \times 8^k + 3f(k)

Reply 2

Gent2324

Careers Forum Helper

Original post by BuryMathsTutor

I wouldn't worry about the way they did it.

You could write

8^{k+1}-3^{k+1}=8 \times 8^k - 3 \times 3^k = 5 \times 8^k + 3 \times 8^k - 3 \times 3^k = 5 \times 8^k + 3f(k)

umm how did u get to the third equation?

Reply 3

RDKGames

Study Forum Helper

Original post by Gent2324

Usually it depends on the expression you want to prove. Here it is

f(k) = 8^n -3^n

so taking away either

8f(k)

3f(k)

would help reduce the problem massively for a beginner.

Though of course, you can prove it from

f(k+1) - f(k)

as well. There are several way go about it and they are all valid.

Taking 8f(k) away:

Spoiler

Taking 3f(k) away:

Spoiler

Taking

f(k)

away:

Spoiler

(edited 5 years ago)

Reply 4

Gent2324

Careers Forum Helper

Original post by BuryMathsTutor

I wouldn't worry about the way they did it.

You could write

8^{k+1}-3^{k+1}=8 \times 8^k - 3 \times 3^k = 5 \times 8^k + 3 \times 8^k - 3 \times 3^k = 5 \times 8^k + 3f(k)

and in the solution bank it says f(k+1) = f(k) + 5(8^k) not plus 3f(k) so im a bit confused on the meaning of it

Reply 5

Gent2324

Careers Forum Helper

Original post by RDKGames

Usually it depends on the expression you want to prove. Here it is

f(k) = 8^n -3^n

so taking away either

8f(k)

3f(k)

would help reduce the problem massively for a beginner.

Though of course, you can prove it from

f(k+1) - f(k)

as well. There are several way go about it and they are all valid.

Taking 8f(k) away:

Spoiler

Taking 3f(k) away:

Spoiler

Taking

f(k)

away:

Spoiler

as for the last one is it assumed that 2f(k) is already divisible by 5? so as a general rule, if i have a term factorised by the number (in this case 5), and then + or - a certain amount of f(k) is it always implied that its divisible by said number? (5)

Reply 6

thotproduct

Here's how I'd go about it

We seek to prove that 8^n - 3^n is a multiple of 5 for all natural numbers >= 1

Base Case:
For n = 1, 8^1 - 3^1 = 5, so it's true for n = 1

Inductive Hypothesis:
If this holds true, for n = some k, where k is a natural number.
8^k - 3^k = 5A, where A is some positive integer

Inductive Step:
Now, if this is true for n = k, we seek to prove that it's true for n = k + 1
We should expect something like 8^(k+1) - 3^(k+1) = 5B, where B is some positive integer

Notice how we can break the indices up, so this is 8 * 8^k - 3 * 3^k = 5B

We can substitute an expression for 8^k from our inductive hypothesis above (for n = k)
5A + 3^k = 8^k
Subbing this in

8(5A + 3^k) - 3* 3^k = 5B
40A + 8*3^k - 3* 3^k = 5B
40A + 5*3^k = 5B
To which you can take out the factor of 5 on the LHS
5(8A + 3^k) = 5B

So you've managed to nab that by factorizing the 5, you've shown it holds true for n = k + 1, given that it's true for n = k, and since it's true for n = 1, the base case, blah blah blah induction conclusion.

You can equivalently do this by the f(k+1) - f(k) method since subtraction will leave you with a multiple of 5, as stated above I wouldn't worry about the way they did it because you can get to an equivalent result with other multiples and other methods.