Find-the-nth-term-100-92-76-44-

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 =108−4×2^n a_n =−(4/3)n^3 +4n^2 −((32)/3)n+108 ...

$${a}_{{n}} =\mathrm{108}−\mathrm{4}×\mathrm{2}^{{n}} \\ $$$${a}_{{n}} =−\frac{\mathrm{4}}{\mathrm{3}}{n}^{\mathrm{3}} +\mathrm{4}{n}^{\mathrm{2}} −\frac{\mathrm{32}}{\mathrm{3}}{n}+\mathrm{108} \\ $$$$… \\ $$

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

100=104−2^2 92=100−2^3 =104−(2^2 +2^3 ) 76=92−2^4 =104−(2^2 +2^3 +2^4 ) ... a_n =104−(2^2 +2^3 +2^4 +...+2^(n+1) ) =104−((2^2 (2^n −1))/(2−1))=108−2^(n+2)

$$\mathrm{100}=\mathrm{104}−\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{92}=\mathrm{100}−\mathrm{2}^{\mathrm{3}} =\mathrm{104}−\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} \right) \\ $$$$\mathrm{76}=\mathrm{92}−\mathrm{2}^{\mathrm{4}} =\mathrm{104}−\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{4}} \right) \\ $$$$… \\ $$$${a}_{{n}} =\mathrm{104}−\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{4}} +…+\mathrm{2}^{{n}+\mathrm{1}} \right) \\ $$$$=\mathrm{104}−\frac{\mathrm{2}^{\mathrm{2}} \left(\mathrm{2}^{{n}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{108}−\mathrm{2}^{{n}+\mathrm{2}} \\ $$

Commented by MJS last updated on 24/Jun/19

we can solve Σ_(i=1) ^4 c_i =100 Σ_(i=1) ^4 c_i ^2 =92 Σ_(i=1) ^4 c_i ^3 =76 Σ_(i=1) ^4 c_i ^4 =44 using c_(1, 2) =α±(√β); c_(3, 4) =γ±(√δ) ⇒ a_n =Σ_(i=1) ^4 c_i ^n but the solution doesn′t look very nice

$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} =\mathrm{100} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{\mathrm{2}} =\mathrm{92} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{\mathrm{3}} =\mathrm{76} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{\mathrm{4}} =\mathrm{44} \\ $$$$\mathrm{using}\:{c}_{\mathrm{1},\:\mathrm{2}} =\alpha\pm\sqrt{\beta};\:{c}_{\mathrm{3},\:\mathrm{4}} =\gamma\pm\sqrt{\delta} \\ $$$$\Rightarrow\:{a}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{c}_{{i}} ^{{n}} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{look}\:\mathrm{very}\:\mathrm{nice} \\ $$

Commented by Cheyboy last updated on 25/Jun/19

$${Thank}\:{you}\:{very}\:{much} \\ $$

Facebook Tweet Pin

Find-the-nth-term-100-92-76-44-

Leave a Reply Cancel reply