Menu Close

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-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *