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# FP1 - [Proof]

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1. So I done the first bit

Step 1
Prove true for n=1 and n = 2

Both true

Step 2
Assume true for n=k and n=k+1, if true for n=k and n=k+1 then true for n=k+2

.....
....
....
Step 1 + Step 2 proves true for all n by induction

I don't know what to do now, may someone help me please
2. You have to show that .
You are correct so far and you can get to the required result from the step you're at.
3. (Original post by B_9710)
You have to show that .
Yeh i know that bit thanks, i just don't know how to make
show
4. (Original post by B_9710)
You have to show that .
You are correct so far and you can get to the required result from the step you're at.
I expanded it but i remember my teacher saying don't expand it unless you absolutely have to, and i can't tell if i have to or not like if there is another option
5. You know you need a , so look for ways to get that. What can you do with 6 and 9?
6. (Original post by NotNotBatman)
You know you need a , so look for ways to get that. What can you do with 6 and 9?
take out a factor of 3?
7. (Original post by AdeptDz)
divide them by 2 and 3? to get 3?
Don't divide, but factorise. Remember you can't change the value.

9 can be written as 3^x right?

And then what happens with the 3?
8. (Original post by NotNotBatman)
Don't divide, but factorise. Remember you can't change the value.

9 can be written as 3^x right?

And then what happens with the 3?
Yeh my bad, i editted the post after you checked it im guessing

ohh, let me try see what i get now
Thanks
9. So would it be
10. (Original post by AdeptDz)
So would it be
Rather than taking factorising 3 in the last step, do you notice that you'll get
in the first term and in the second term there is a remembering that

Also you've written it a bit wrong. It is

remember it's 3^k, not 3k, so you can't factorise 3 like that.
11. (Original post by NotNotBatman)
Rather than taking factorising 3 in the last step, do you notice that you'll get
in the first term and in the second term there is a remembering that

Also you've written it a bit wrong. It is

remember it's 3^k, not 3k, so you can't factorise 3 like that.
cool so:
12. (Original post by AdeptDz)
cool so:
The adding power rule only applies when it's the same base, so

So you want on the first term; , so the powers of 3 can be added. do the same with

But leave the brackets (k-1) and (k-2) for now.
13. (Original post by NotNotBatman)
The adding power rule only applies when it's the same base, so

So you want on the first term; , so the powers of 3 can be added. do the same with

But leave the brackets (k-1) and (k-2) for now.
Would it be

so
14. (Original post by AdeptDz)
Would it be

so
yes, then you can add the powers of 3 on each individual term.
15. (Original post by AdeptDz)
Would it be

so
so
16. (Original post by AdeptDz)
so
Yes, then you can factorise.
17. (Original post by NotNotBatman)
Yes, then you can factorise.

Got a feeling i did that wrong as idk where to put the 2
18. (Original post by AdeptDz)

Got a feeling i did that wrong as idk where to put the 2
Factorise the whole thing, so use a big bracket and leave the 2 on the first term, it doesn't matter which way around it's multiplied.

when you multiply out it's the same right?
19. (Original post by AdeptDz)
so
(Original post by NotNotBatman)
Yes, then you can factorise.
1sec, let me have another go

20. there

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