Question 120832
Assume you mean:
{{{4/(3n)}}} + {{{2/((n+1))}}} + {{{2/((n^2+n))}}}
:
Factor out n in the 3rd denominator:
{{{4/(3n)}}} + {{{2/((n+1))}}} + {{{2/(n(n+1))}}}
:
The common denominator will be: 3n(n+1), so we have:
:
{{{(4(n+1) + 2(3n) + 3(2))/(3n(n+1))}}} = {{{(4n + 4 + 6n + 6)/(3n(n+1))}}} = {{{(10n + 10)/(3n(n+1))}}} = {{{(10(n + 1))/(3n(n+1))}}}
:
The (n+1)'s cancel, we are left with:
{{{10/(3n)}}}
:
How about this? Could you follow this ok? Any questions? Let me know.