Question 76767
(4/3n)+(2/n+1)+(2/n^2+n)
:
{{{4/(3n)+2/(n+1)+2/(n^2+n)}}}
:
{{{4/(3n)+2/(n+1)+2/(n(n+1))}}}
:
The common denominator would be 3n(n+1):
{{{(4(n+1) + 3n(2) + 3(2))/(3n(n+1))}}}
:
{{{(4n + 4 + 6n + 6)/(3n(n+1))}}}
:
{{{(10n + 10)/(3n(n+1))}}} =  {{{(10(n + 1))/(3n(n+1))}}} = {{{10/(3n)}}}
:
Did all this make sense to you?