Question 1136725
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The equation is not true for any value of n.  If the equation is supposed to have a solution, then you have not shown the equation correctly.<br>
(n-1)C(n-3) is the same as (n-1)C(2); (n+1)C(n-2) is the same as (n+1)C(3).  The equation is then<br>
{{{4((n-1)(n-2)/2) = 3((n+1)(n)(n-1)/6)}}}
{{{2(n-2) = (n(n+1))/2}}}
{{{4n-8 = n^2+n}}}
{{{n^2-3n+8 = 0}}}<br>
The solutions are not real numbers....