• Nov 2nd 2005, 03:30 PM
Natasha
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 PM
eah1010
ummm...
The only thing i got was

5/2n^2 + 8n + 7/4
• Nov 2nd 2005, 04:51 PM
Jameson
Quote:

Originally Posted by Natasha
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)

Ok. So you have this equation:

$\frac{n^2+(n+1)^2}{4}+3\left(\frac{n(n+1)(2n+1)}{6 }\right)+2\left(\frac{n(n+1)}{2}\right)$

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 PM
Natasha
Right 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 AM
Natasha
((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 AM
CaptainBlack
Quote:

Originally Posted by Natasha
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)

I suspect that you may have an typo in one or other
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 AM
Natasha
You are spot on. I had made a typo mistake... Resolve it now thanks ;)