# From point A to point B, basic algebra.

1. Jan 14, 2009

### calisoca

1. The problem statement, all variables and given/known data

I'm not sure how to get from point A to point B. It seems simple enough, but I'm just not seeing it!

2. Relevant equations

point A:

$$\frac{81}{n^4} \ [\frac{n(n+1)}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n(n+1)}{2}]$$

point B:

$$\frac{81}{4} \ (1 + \frac{1}{n})^2 \ - \ 27(1 + \frac{1}{n})$$

3. The attempt at a solution

I'm honestly at a loss on this simple problem, however sad that may seem.

Last edited: Jan 14, 2009
2. Jan 14, 2009

### danago

For the left hand term, distribute the power of 2 within the bracket and make any obvious cancellations. Then use the fact that an/bn = (a/b)n to get it closer to the desired form. See if you can figure out the last step.

For the right hand term, first make an obvious cancellation, and then break the fraction up as required.

I have tried to only give hints so you can do the actual math yourself. Hopefully one of the hints may be the thing you were looking for and you can finish it off yourself

3. Jan 14, 2009

### calisoca

Danago, thank you very much. I will work with the hints you gave me and see if I can figure it out. It's not hard, it's just that I haven't actually done any real algebra in the last few months, so I'm a bit slow right now. If I need any more help from here, I'll post, but otherwise, thank you again for your help!

4. Jan 14, 2009

### calisoca

Okay, here is what I have so far. However, getting the first term to what I want is problematic for me. Step 6 is where I am stuck. May I please get some help getting past Step 6? I'd greatly appreciate it.

1.) $$\frac{81}{n^4} \ [\frac{n(n+1)}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n(n+1)}{2}]$$

2.) $$\frac{81}{n^4} \ [\frac{n^2 + n}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n^2 + n}{2}]$$

3.) $$\frac{81}{n^4} \ [\frac{n^4 + 2n^3 + n^2}{4}] \ - \ \frac{54}{n^2} \ [\frac{n^2 + n}{2}]$$

4.) $$\frac{81(n^4 + 2n^3 + n^2)}{4n^4} \ - \ \frac{54(n^2 + n)}{2n^2}$$

5.) $$\frac{81}{4} \ [\frac{n^4 + 2n^3 + n^2}{n^4}] \ - \ \frac{54}{2} \ [\frac{n^2 + n}{n^2}]$$

6.) $$\frac{81}{4} \ (1 + \frac{2}{n} + \frac{1}{n^2}}) \ - \ \frac{54}{2} \ (1 + \frac{1}{n})$$

5. Jan 14, 2009

### calisoca

Crap! Just figured it out. Like Danago said, $$\frac{a^2}{b^2} = (\frac{a}{b})^2$$

So....

1.) $$\frac{81}{n^4} \ [\frac{n(n+1)}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n(n+1)}{2}]$$

2.) $$\frac{81}{n^4} \ [\frac{n^2 + n}{2}]^2 \ - \ \frac{54}{n^2} \ [\frac{n^2 + n}{2}]$$

3.) $$\frac{81}{n^4} \ [\frac{n^4 + 2n^3 + n^2}{4}] \ - \ \frac{54}{n^2} \ [\frac{n^2 + n}{2}]$$

4.) $$\frac{81(n^4 + 2n^3 + n^2)}{4n^4} \ - \ \frac{54(n^2 + n)}{2n^2}$$

5.) $$\frac{81}{4} \ [\frac{n^4 + 2n^3 + n^2}{n^4}] \ - \ \frac{54}{2} \ [\frac{n^2 + n}{n^2}]$$

6.) $$\frac{81}{4} \ (1 + \frac{2}{n} + \frac{1}{n^2}}) \ - \ \frac{54}{2} \ (1 + \frac{1}{n})$$

7.) $$(1 + \frac{2}{n} + \frac{1}{n^2}) = (1 + \frac{1}{n})^2$$

8.) $$\frac{81}{4} \ (1 + \frac{1}{n})^2 \ - \ 27(1 + \frac{1}{n})$$

Thanks for the help!

6. Jan 14, 2009

### danago

No problems You could actually have done it in a slightly simpler way, without expanding the brackets how you did. Ill demonstrate with the left hand term:

$$\frac{81}{n^4} \ [\frac{n(n+1)}{2}]^2 =\frac{81}{n^4} \ [\frac{n^2(n+1)^2}{4}] =\frac{81}{4} \ [\frac{(n+1)^2}{n^2}] =\frac{81}{4} \ [\frac{(n+1)}{n}]^2 =\frac{81}{4} \ [1+\frac{1}{n}]^2$$

It still ends with the same result, but its probably a little simpler