Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 12267 by tawa last updated on 17/Apr/17

Find the nth term of the sequence 1) (1/3) , (1/(15)) , (1/(35)) , (1/(63)) , (1/(99)) 2) (1/2), (1/6), (1/(12)), (1/(20)), (1/(30))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence} \\ $$$$\left.\mathrm{1}\right)\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\:,\:\frac{\mathrm{1}}{\mathrm{15}}\:,\:\frac{\mathrm{1}}{\mathrm{35}}\:,\:\frac{\mathrm{1}}{\mathrm{63}}\:,\:\frac{\mathrm{1}}{\mathrm{99}} \\ $$$$\left.\mathrm{2}\right)\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{6}},\:\frac{\mathrm{1}}{\mathrm{12}},\:\frac{\mathrm{1}}{\mathrm{20}},\:\frac{\mathrm{1}}{\mathrm{30}} \\ $$

Answered by ajfour last updated on 17/Apr/17

1) T_n =(1/(4n^2 −1)) 2) T_n =(1/(n^2 +n))

$$\left.\mathrm{1}\right)\:\:{T}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:\:{T}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}} \\ $$

Commented by tawa last updated on 17/Apr/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Joel576 last updated on 17/Apr/17

1) U_1 = (1/3) = (1/(1 . 3)) U_2 = (1/(15)) = (1/(3 . 5)) U_3 = (1/(35)) = (1/(5 . 7)) U_n = (1/((2n − 1)(2n + 1))) = (1/(4n^2 − 1)) 2) U_1 = (1/2) = (1/(1 . 2)) U_2 = (1/6) = (1/(2 . 3)) U_3 = (1/(12)) = (1/(3 . 4)) U_n = (1/(n(n + 1))) = (1/(n^2 + n))

$$\left.\mathrm{1}\right)\:{U}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:.\:\mathrm{3}} \\ $$$$\:\:\:\:\:\:{U}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{15}}\:=\:\frac{\mathrm{1}}{\mathrm{3}\:.\:\mathrm{5}} \\ $$$$\:\:\:\:\:\:{U}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{35}}\:=\:\frac{\mathrm{1}}{\mathrm{5}\:.\:\mathrm{7}} \\ $$$$\:\:\:\:\:\:{U}_{{n}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\:−\:\mathrm{1}\right)\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{U}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:.\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:{U}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\:.\:\mathrm{3}}\:\:\:\: \\ $$$$\:\:\:\:\:\:{U}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{12}}\:=\:\frac{\mathrm{1}}{\mathrm{3}\:.\:\mathrm{4}} \\ $$$$\:\:\:\:\:\:{U}_{{n}} \:=\:\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\:{n}} \\ $$

Commented by tawa last updated on 17/Apr/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com