Question Number 25112 by Mr easy last updated on 04/Dec/17
Commented by prakash jain last updated on 04/Dec/17
$$\left({a}+{b}\right)^{\mathrm{2}{n}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} {a}^{{i}} {b}^{\mathrm{2}{n}−{i}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} \left(\mathrm{2}\right)^{{i}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}−{i}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} \left(\mathrm{2}\right)^{{i}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}−{i}} \left(−\mathrm{1}\right)^{\mathrm{2}{n}−{i}} \\ $$$$\left(\mathrm{1}\right)^{\mathrm{2n}−{i}} \:\mathrm{is}\:\left(−\mathrm{1}\right)\:\mathrm{for}\:\mathrm{odd}\:\mathrm{terms}. \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} −\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} \left(\mathrm{2}\right)^{{i}} \left(\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}−{i}} +\left(−\mathrm{1}\right)^{\mathrm{2}{n}−{i}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}−{i}} \right) \\ $$$$\left.=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{\mathrm{2}{i}} \left(\mathrm{2}\right)^{\mathrm{2}{i}} \left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}−\mathrm{2}{i}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}−\mathrm{2}{i}} \right) \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{\mathrm{2}{i}} \left(\mathrm{2}\right)^{\mathrm{2}{i}} \left(\mathrm{3}\right)^{{n}−{i}} ×\mathrm{2}\:\:\left(\mathrm{an}\:\mathrm{even}\:\mathrm{integer}\right) \\ $$$$\because\:{i}\:\mathrm{odd}\:\mathrm{term}\:\mathrm{will}\:\mathrm{cancel}. \\ $$