Question Number 132961 by danielasebhofoh last updated on 17/Feb/21
Answered by Ar Brandon last updated on 17/Feb/21
$$\mathrm{2n}−\mathrm{3}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{3}} {\sum}}\mathrm{k}=\mathrm{2n}−\mathrm{3}+\frac{\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{4n}−\mathrm{6}+\mathrm{n}^{\mathrm{2}} −\mathrm{5n}+\mathrm{6}}{\mathrm{2}}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{2}} =\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{2}\right)!\mathrm{2}!}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$