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If-u-1-u-2-u-3-u-n-2n-2-n-is-an-AP-Find-u-1-u-2-u-3-u-2n-2-u-2n-1-

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Question Number 128369 by I want to learn more last updated on 06/Jan/21
If u_1 + u_2 + u_3 + ... + u_n = 2n^2 + n is an AP. Find u_1 + u_2 + u_3 + ... + u_(2n − 2) + u_(2n − 1)
$$\mathrm{If}\:\:\:\mathrm{u}_{\mathrm{1}} \:\:+\:\:\mathrm{u}_{\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{3}} \:\:+\:\:…\:\:+\:\:\mathrm{u}_{\mathrm{n}} \:\:\:=\:\:\:\mathrm{2n}^{\mathrm{2}} \:\:+\:\:\mathrm{n}\:\:\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}. \\ $$$$\mathrm{Find}\:\:\:\:\:\:\:\:\mathrm{u}_{\mathrm{1}} \:\:+\:\:\mathrm{u}_{\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{3}} \:\:+\:\:…\:\:+\:\:\mathrm{u}_{\mathrm{2n}\:\:−\:\:\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{2n}\:\:−\:\:\mathrm{1}} \\ $$
Answered by mr W last updated on 06/Jan/21
u_1 + u_2 + u_3 + ... + u_(2n − 2) + u_(2n − 1) =2(2n−1)^2 +(2n−1) =(2n−1)(4n−1) =8n^2 −6n+1
$$\mathrm{u}_{\mathrm{1}} \:\:+\:\:\mathrm{u}_{\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{3}} \:\:+\:\:…\:\:+\:\:\mathrm{u}_{\mathrm{2n}\:\:−\:\:\mathrm{2}} \:\:+\:\:\mathrm{u}_{\mathrm{2n}\:\:−\:\:\mathrm{1}} \\ $$$$=\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$=\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{4}{n}−\mathrm{1}\right) \\ $$$$=\mathrm{8}{n}^{\mathrm{2}} −\mathrm{6}{n}+\mathrm{1} \\ $$
Commented by I want to learn more last updated on 06/Jan/21
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 06/Jan/21
we have u_1 +u_2 +...+u_n =2n^(2 ) +n ⇒ { ((u_1 +u_2 +...+u_n =2n^2 +n)),((u_1 +u_2 +..u_(n−1) =2(n−1)^2 +n−1 ⇒)) :} u_n =2n^2 +n−2(n−1)^2 −n+1 =2n^2 −2(n^2 −2n+1)+1 =4n−2 +1 =4n−1 so u_1 +u_2 +....+u_(2n−1) =((2n−1−1+1)/2)(u_1 +u_(2n−1) ) =((2n−1)/2)(3 +4(2n−1)−1) =((2n−1)/2)(3+8n−4−1) =(((2n−1)(8n−2))/2) =(2n−1)(4n−1) =8n^2 −2n−4n+1 =8n^2 −6n +1
$$\mathrm{we}\:\mathrm{have}\:\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} +…+\mathrm{u}_{\mathrm{n}} =\mathrm{2n}^{\mathrm{2}\:} +\mathrm{n}\:\Rightarrow \\ $$$$\begin{cases}{\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} +…+\mathrm{u}_{\mathrm{n}} =\mathrm{2n}^{\mathrm{2}} \:+\mathrm{n}}\\{\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} +..\mathrm{u}_{\mathrm{n}−\mathrm{1}} =\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{n}−\mathrm{1}\:\Rightarrow}\end{cases} \\ $$$$\mathrm{u}_{\mathrm{n}} =\mathrm{2n}^{\mathrm{2}} +\mathrm{n}−\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{n}+\mathrm{1}\:=\mathrm{2n}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{1}\right)+\mathrm{1} \\ $$$$=\mathrm{4n}−\mathrm{2}\:+\mathrm{1}\:=\mathrm{4n}−\mathrm{1}\:\mathrm{so} \\ $$$$\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} +….+\mathrm{u}_{\mathrm{2n}−\mathrm{1}} =\frac{\mathrm{2n}−\mathrm{1}−\mathrm{1}+\mathrm{1}}{\mathrm{2}}\left(\mathrm{u}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2n}−\mathrm{1}} \right) \\ $$$$=\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}\:+\mathrm{4}\left(\mathrm{2n}−\mathrm{1}\right)−\mathrm{1}\right)\:=\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}+\mathrm{8n}−\mathrm{4}−\mathrm{1}\right) \\ $$$$=\frac{\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{8n}−\mathrm{2}\right)}{\mathrm{2}}\:=\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{4n}−\mathrm{1}\right) \\ $$$$=\mathrm{8n}^{\mathrm{2}} −\mathrm{2n}−\mathrm{4n}+\mathrm{1}\:=\mathrm{8n}^{\mathrm{2}} −\mathrm{6n}\:+\mathrm{1} \\ $$
Commented by I want to learn more last updated on 07/Jan/21
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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