Can a Sequence be Simplified Using Exponent Rules?

[nobahar]

Hi guys,
the x's are supposed to be *, as in 2x2 is supposed to be 2*2, it won't listen to me.
I was wondering if it is possible to simplify this sequence:

1)2

2)2^2 +4

3)2^3 +(2 * 4) + 6

4)2^4 + (2^2 * 4) + (2 * 6) +8

5)2^5 + (2^3 * 4) + (2^2 * 6) + (2 * 8) + 10

6)2^6 + (2^4 * 4) + (2^3 * 6) + (2^2 * 8) + (2 * 10) + 12

I've made a few attempts but there are not satisfactory:
2^n + 2^n-2 * (n-(n-2))+2^n-3 * (n-(n-4))+...+2^2 * (2n-2) + 2^0 * 2n
Any help appreciated

 

[chroot]

It's pretty easy to do if you're able to use functions like max().

- Warren

 

[nobahar]

Yeah, my attempt in Latex code didn't look how it was supposed to, I'm not particularly good with Latex. Thanks for the advice, I'll go look up Max().

 

[nobahar]

The only maximum function my maths textbook gives me is for trig and for calculus. Maximum values in trig don't apply here, and I can't see the application of the maximum in calculus here either. Any further advice, hints, or tips?

 

[Mark44]

Hi guys,
the x's are supposed to be *, as in 2x2 is supposed to be 2*2, it won't listen to me.
I was wondering if it is possible to simplify this sequence:

1)2

2)2^2 +4

3)2^3 +(2 * 4) + 6

4)2^4 + (2^2 * 4) + (2 * 6) +8

5)2^5 + (2^3 * 4) + (2^2 * 6) + (2 * 8) + 10

6)2^6 + (2^4 * 4) + (2^3 * 6) + (2^2 * 8) + (2 * 10) + 12

I've made a few attempts but there are not satisfactory:
2^n + 2^n-2 * (n-(n-2))+2^n-3 * (n-(n-4))+...+2^2 * (2n-2) + 2^0 * 2n
Any help appreciated

It looks like you're on the right track. The nth term of your sequence will have n terms in it, with the first term being 2^n. The other n - 1 terms seem to follow their own pattern.

In each of the remaining n - 1 terms, you have the product of 2 to some power and twice a number. The exponent on 2 starts at n - 2 and works its way down to 0. The "twice a number" factor starts at 2*2 and works its way up to 2*n.

Does that help?

 

[nobahar]

Thanks so much for your help Mark.
Okay, I think this is right, after some further simplification:
2^n+2\sum_{a=2}^n a2^{n-a}

 

[Mark44]

Can I assume that this is what you're looking for? Your comment about being stuck at n = 6 made me unsure.

 

[nobahar]

Sorry, the equation I posted with 'n' is the one I wanted. Previously I entered 6 in it's place accidently (I was reading it from my working and used one where I subtituted n with 6 to test the equation) and when I edited the Latex in the post it didn't change from 6 to n (I assumed there would be a delay of 24 hours or soemthing), but it has now. If that makes sense!
But yes, the equation, as it appears above (with 'n'), is it correct?

 

[Mark44]

It looks like what I described. If your formula gives you the right values for the numbers in your sequence, I think all is good!

 

[nobahar]

Thanks Mark: I tried it with two (an exhaustive test), and it appears to work. It is certainly simplified! Thankyou.