Question Number 62664 by Cheyboy last updated on 24/Jun/19
$${Find}\:{the}\:{nth}\:{term} \\ $$$$\mathrm{100},\mathrm{92},\mathrm{76},\mathrm{44},…… \\ $$
Answered by MJS last updated on 24/Jun/19
$${a}_{{n}} =\mathrm{108}−\mathrm{4}×\mathrm{2}^{{n}} \\ $$$${a}_{{n}} =−\frac{\mathrm{4}}{\mathrm{3}}{n}^{\mathrm{3}} +\mathrm{4}{n}^{\mathrm{2}} −\frac{\mathrm{32}}{\mathrm{3}}{n}+\mathrm{108} \\ $$$$… \\ $$
Commented by ajfour last updated on 24/Jun/19
wonderful sir! how to expect and judge the first one?
Commented by mr W last updated on 24/Jun/19
$$\mathrm{100}=\mathrm{104}−\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{92}=\mathrm{100}−\mathrm{2}^{\mathrm{3}} =\mathrm{104}−\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} \right) \\ $$$$\mathrm{76}=\mathrm{92}−\mathrm{2}^{\mathrm{4}} =\mathrm{104}−\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{4}} \right) \\ $$$$… \\ $$$${a}_{{n}} =\mathrm{104}−\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{4}} +…+\mathrm{2}^{{n}+\mathrm{1}} \right) \\ $$$$=\mathrm{104}−\frac{\mathrm{2}^{\mathrm{2}} \left(\mathrm{2}^{{n}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{108}−\mathrm{2}^{{n}+\mathrm{2}} \\ $$
Commented by MJS last updated on 24/Jun/19
$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} =\mathrm{100} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{\mathrm{2}} =\mathrm{92} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{\mathrm{3}} =\mathrm{76} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{\mathrm{4}} =\mathrm{44} \\ $$$$\mathrm{using}\:{c}_{\mathrm{1},\:\mathrm{2}} =\alpha\pm\sqrt{\beta};\:{c}_{\mathrm{3},\:\mathrm{4}} =\gamma\pm\sqrt{\delta} \\ $$$$\Rightarrow\:{a}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{{n}} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{look}\:\mathrm{very}\:\mathrm{nice} \\ $$
Commented by Cheyboy last updated on 25/Jun/19
$${Thank}\:{you}\:{very}\:{much} \\ $$