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If-u-1-u-2-u-3-u-n-2n-2-n-is-a-AP-find-the-value-of-u-1-u-2-u-3-u-2n-2-u-2n-1-

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Question Number 83406 by jagoll last updated on 02/Mar/20
If u_1 +u_2 +u_3 +...+u_n = 2n^2 +n is a AP. find the value of u_1 +u_2 +u_3 +...+u_(2n−2) +u_(2n−1) .
Ifu1+u2+u3+…+un=2n2+nisaAP.findthevalueofu1+u2+u3+…+u2n−2+u2n−1.
Commented by jagoll last updated on 03/Mar/20
my way from s_n = 2n^2 +n { ((u_1 =3)),((u_n = 4n−1)) :} u_1 +u_2 +u_3 +...+u_(2n−2) +u_(2n−1) = M let 2n−1 = k M = (k/2)(3+u_k )=(k/2)(3+4k−1) M = (k/2)(4k+2) = k (2k+1) M = (2n−1)(4k−1) = 8n^2 −6n+1
mywayfromsn=2n2+n{u1=3un=4n−1u1+u2+u3+…+u2n−2+u2n−1=Mlet2n−1=kM=k2(3+uk)=k2(3+4k−1)M=k2(4k+2)=k(2k+1)M=(2n−1)(4k−1)=8n2−6n+1
Commented by mathmax by abdo last updated on 02/Mar/20
we have u_1 +u_2 +....+u_n =2n^2 +n u_1 +u_2 +....+u_n +u_(n+1) =2(n+1)^2 +n+1 ⇒ u_(n+1) =2(n+1)^2 +n+1−2n^2 −n =2(n^2 +2n+1)−2n^2 +1 =2n^2 +4n+2−2n^2 +1 =4n+3 ⇒u_n =4(n−1)+3 =4n−1 ⇒ 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)(2+8n−4) =(2n−1)(4n−1)=8n^2 −2n−4n+1 =8n^2 −6n +1
wehaveu1+u2+….+un=2n2+nu1+u2+….+un+un+1=2(n+1)2+n+1⇒un+1=2(n+1)2+n+1−2n2−n=2(n2+2n+1)−2n2+1=2n2+4n+2−2n2+1=4n+3⇒un=4(n−1)+3=4n−1⇒u1+u2+….+u2n−1=2n−1−1+12(u1+u2n−1)=2n−12{3+4(2n−1)−1}=2n−12(2+8n−4)=(2n−1)(4n−1)=8n2−2n−4n+1=8n2−6n+1
Commented by mathmax by abdo last updated on 02/Mar/20
u_n =4n−1
un=4n−1
Answered by mind is power last updated on 02/Mar/20
ΣU_k =(((u_1 +u_n )n)/2) u_n =U_1 +(n−1)r ⇒(((2u_1 +(n−1)r)n)/2)=2n^2 +n ⇒2u_1 +(n−1)r=4n+2 2u_1 −r+nr=4n+2 ⇒r=4,u_1 =3 Σ_(k=1) ^(2n−1) u_k =((2n−1)/2)(u_1 +u_(2n−1) ) =((2n−1)/2)(u_1 +u_1 +(2n−2)r) =((2n−1)/2)(6+(2n−2)4)= =(2n−1)(−1+4n)=8n^2 −6n+1
ΣUk=(u1+un)n2un=U1+(n−1)r⇒(2u1+(n−1)r)n2=2n2+n⇒2u1+(n−1)r=4n+22u1−r+nr=4n+2⇒r=4,u1=3∑2n−1k=1uk=2n−12(u1+u2n−1)=2n−12(u1+u1+(2n−2)r)=2n−12(6+(2n−2)4)==(2n−1)(−1+4n)=8n2−6n+1
Commented by mind is power last updated on 02/Mar/20
2n−1=11 sir
2n−1=11sir
Commented by jagoll last updated on 02/Mar/20
if n = 6⇒ 8(36)−36+1= 7×36+1=253 u_1 +u_2 +u_3 +...+u_(10 ) = 5(3+39) = 5×42 =210 not same sir
ifn=6⇒8(36)−36+1=7×36+1=253u1+u2+u3+…+u10=5(3+39)=5×42=210notsamesir
Answered by mr W last updated on 02/Mar/20
S_n =2n^2 +n=n(2n+1) S_(2n−1) =(2n−1)(2(2n−1)+1)=(2n−1)(4n−1)
Sn=2n2+n=n(2n+1)S2n−1=(2n−1)(2(2n−1)+1)=(2n−1)(4n−1)
Commented by mr W last updated on 02/Mar/20
if we want to know u_n , then u_n =S_n −S_(n−1) =2n^2 +n−2(n−1)^2 −(n−1) =4n−1
ifwewanttoknowun,thenun=Sn−Sn−1=2n2+n−2(n−1)2−(n−1)=4n−1
Answered by TANMAY PANACEA last updated on 02/Mar/20
put n=1 u_1 =3 put n=2 u_1 +u_2 =10 so u_2 =7 n=3 u_1 +u_2 +u_3 =21 so u_3 =11 so first term ofA.P=3 common difference=4 S_(2n−1) =((2n−1)/2)[2×3+(2n−1−1)×4] =((2n−1)/2)×[6+8n−8] =(2n−1)(4n−1)
putn=1u1=3putn=2u1+u2=10sou2=7n=3u1+u2+u3=21sou3=11sofirsttermofA.P=3commondifference=4S2n−1=2n−12[2×3+(2n−1−1)×4]=2n−12×[6+8n−8]=(2n−1)(4n−1)

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