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:Originally Posted by Natasha
$\displaystyle \frac{n^2+(n+1)^2}{4}+3\left(\frac{n(n+1)(2n+1)}{6 }\right)+2\left(\frac{n(n+1)}{2}\right)$
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
((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 :-)
((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 :-)