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(n+1)(n-1) Indices

1. Hi, this may be sixth form, not sure.

How does this work, cannot quite fathom it, no help in book so am left in ambiguity.

So, I am supposed to show that , for

fine

fine

Now, what I don't get is how this line becomes the next

How does that work?
2. They factorised it using the laws of indices.
3. (Original post by SubAtomic)
Hi, this may be sixth form, not sure.

How does this work, cannot quite fathom it, no help in book so am left in ambiguity.

So, I am supposed to show that , for

fine

fine

Now, what I don't get is how this line becomes the next

How does that work?
Just notice that , similar thing for the others.
4. (Original post by anil10100)
Just notice that , similar thing for the others.
Nice, will give it a whirl and see where I end up.
5. So are these equivalent then

Seems to be according to the calculator.

Back to the question, so

Is this right? What now? How do I get rid of those n in the bracket so it looks like the last bit of my OP?
6. (Original post by SubAtomic)
So are these equivalent then

Seems to be according to the calculator.
In general,

7. (Original post by raheem94)
In general,

Yeah that was my own little spin on things, not too sure on the second bit of that post.
8. (Original post by SubAtomic)
Yeah that was my own little spin on things, not too sure on the second bit of that post.
Spoiler:
Show

9. (Original post by SubAtomic)
Now, what I don't get is how this line becomes the next

How does that work?

For easier comprehension, let's break it down by isolating the in each individual term.

First term:

Spoiler:
Show
Sub in n+1=n-1+2:

Apply the rule a^(x+y)=(a^x)(a^y):

Factorize 3^(n-1)*4^(n-1) out (although it already is factorized because it's all multiplication here):

Second term:

Spoiler:
Show
Sub in n+1=(n-1)+2:

Apply the rule a^(x+y)=(a^x)(a^y):

Factorize 3^(n-1)*4^(n-1) out (although it already is factorized because it's all multiplication here):

Third term:

Spoiler:
Show
Sub in n=(n-1)+1:

Apply the rule a^(x+y)=(a^x)(a^y)

Factorize 3^(n-1)*4^(n-1) out (although it already is factorized because it's all multiplication here):

All together now:

Factorize out 3^(n-1)*4^(n-1):

10. (Original post by aznkid66)
...
Nice explanation again, I thought of subbing the n=(n-1)+1 into the but thought it a bit of an odd way of doing things for some reason you took it that extra step, thanks. You are missing the closing brackets in your third spoiler but I get ya

(Original post by raheem94)
...
Nice one mate, both your explanations have helped.

Must take quite a while to be able to rep people again
11. So let me see if I have got this right by trying another, tried one earlier and it went horribly wrong, I think.

This first

for

Can I do something with 2^2 in the indices here? Seen as 4 and 8 are 2^2 and 2^3. Or would I be over-complicating things?

Will carry on as per anyway,

So

Is this right? If it is can I simplify further?
12. (Original post by SubAtomic)
So let me see if I have got this right by trying another, tried one earlier and it went horribly wrong, I think.

This first

for

Can I do something with 2^2 in the indices here? Seen as 4 and 8 are 2^2 and 2^3. Or would I be over-complicating things?

13. (Original post by raheem94)
...
Yep I understand this, was meaning something like

Just trying stuff nothing major, seeing if a neater way.
14. (Original post by SubAtomic)
Yep I understand this, was meaning something like

Just trying stuff nothing major, seeing if a neater way.
15. (Original post by SubAtomic)
This is wrong.

If you want to check it, sub in a number.

16. (Original post by raheem94)
..
Edited, experiments and all.
17. (Original post by raheem94)
...
The main thing I was wondering about is how the indices have common factors and if they can be simplified in a certain fashion, not actually quite sure what I was thinking but it was something lol.

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