How can I go from

((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

to this

(1/4)n(n+1)(n+2)(n+3)

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- Nov 2nd 2005, 03:30 PMNatashaurgent please algebra solving...
How can I go from

((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

to this

(1/4)n(n+1)(n+2)(n+3) - Nov 2nd 2005, 03:50 PMeah1010ummm...
The only thing i got was

5/2n^2 + 8n + 7/4 - Nov 2nd 2005, 04:51 PMJamesonQuote:

Originally Posted by**Natasha**

I would suggest getting a common denominator. Then expand things out, and simplify. There may be a more elegant way to do it, but it won't hurt to practice algebra.

Jameson - Nov 3rd 2005, 12:24 PMNatashaRight then...
((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

((12n^3+36n^2+14n+1)) / 12

And I need to get

(1/4)n(n+1)(n+2)(n+3)

Can someone help the simple further steps to take thanks :-) - Nov 4th 2005, 04:29 AMNatasha
((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

**((4n^3+12n^2+8n+1)) / 4**

And I need to get

(1/4)n(n+1)(n+2)(n+3)

Can someone help the simple further steps to take thanks :-) - Nov 9th 2005, 10:06 AMCaptainBlackQuote:

Originally Posted by**Natasha**

of these expressions.

Let n=0, then the first simplifies to 1/4,

while the second simplies to 0.

RonL - Nov 9th 2005, 11:08 AMNatasha
You are spot on. I had made a typo mistake... Resolve it now thanks ;)