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Question-7102




Question Number 7102 by Tawakalitu. last updated on 10/Aug/16
Commented by 123456 last updated on 10/Aug/16
S_0 =2  S_1 =2
S0=2S1=2
Commented by prakash jain last updated on 10/Aug/16
x+y=2  xy=4  (x+y)^2 −4xy=(x−y)^2 =4−16=−12  x−y=2(√3)i  x=1+i(√3)=2e^(iπ/3)   y=1−i(√3)=2e^(−iπ/3)   x^n +y^n =2^n e^(inπ/3) +2^n e^(−inπ/3) =2^(n+1) cos ((nπ)/3)  S_(n+1) =2^(n+2) cos (((n+1)π)/3)  S_(n−1) =2^n cos (((n−1)π)/3)  S_(n+1) +4S_(n−1) =2^(n+2) [cos (((n+1)π)/3)+cos (((n−1)π)/3)]  =2^(n+2) [2cos ((nπ)/3)cos (π/3)]=2^(n+3) cos ((nπ)/3)=4[2^(n+1) cos ((nπ)/3)]  =2[2^(n+1) cos ((nπ)/3)×(1/2)]=2S_n
x+y=2xy=4(x+y)24xy=(xy)2=416=12xy=23ix=1+i3=2eiπ/3y=1i3=2eiπ/3xn+yn=2neinπ/3+2neinπ/3=2n+1cosnπ3Sn+1=2n+2cos(n+1)π3Sn1=2ncos(n1)π3Sn+1+4Sn1=2n+2[cos(n+1)π3+cos(n1)π3]=2n+2[2cosnπ3cosπ3]=2n+3cosnπ3=4[2n+1cosnπ3]=2[2n+1cosnπ3×12]=2Sn
Commented by Tawakalitu. last updated on 10/Aug/16
I really appreciate your effort, thank you so much
Ireallyappreciateyoureffort,thankyousomuch
Commented by Tawakalitu. last updated on 10/Aug/16
Workings please
Workingsplease
Commented by Tawakalitu. last updated on 10/Aug/16
please edit your wotkings and show how p = 4, q = 4. thanks
pleaseedityourwotkingsandshowhowp=4,q=4.thanks
Commented by Yozzii last updated on 10/Aug/16
Check your second to last line of text.  2^(n+2) [2cos((nπ)/3)cos(π/3)]=2^(n+2) [2cos((nπ)/3)×(1/2)]=2×2^(n+1) cos((nπ)/3)  ⇒p=2 ∴ pq=8
Checkyoursecondtolastlineoftext.2n+2[2cosnπ3cosπ3]=2n+2[2cosnπ3×12]=2×2n+1cosnπ3p=2pq=8
Commented by prakash jain last updated on 10/Aug/16
Thanks. Corrected.
Thanks.Corrected.
Commented by prakash jain last updated on 10/Aug/16
Please see last three lines for relation  between S_(n+1) , S_(n−1)  and S_n
PleaseseelastthreelinesforrelationbetweenSn+1,Sn1andSn
Commented by Tawakalitu. last updated on 10/Aug/16
Seen thanks so much, i can understand now
Seenthankssomuch,icanunderstandnow
Answered by prakash jain last updated on 10/Aug/16
p=2  q=4  pq=16
p=2q=4pq=16
Commented by Tawakalitu. last updated on 10/Aug/16
Workings please
Workingsplease
Commented by Yozzii last updated on 10/Aug/16
p=2, q=4, pq=8
p=2,q=4,pq=8
Answered by nume1114 last updated on 12/Aug/16
 { ((x+y=2)),((xy=4)),((S_n =x^n +y^n )) :}  S_(n+1) =x^(n+1) +y^(n+1)             =(x+y)(x^n +y^n )−xy(x^(n−1) +y^(n−1) )            (((    (x+y)(x^n +y^n ))),((=x^(n+1) +y^(n+1) +x^n y+xy^n )),((=x^(n+1) +y^(n+1) +xy(x^(n−1) +y^(n−1) ))) )            =(x+y)S_n −xyS_(n−1)             =2S_n −4S_(n−1)     2S_n =S_(n+1) +4S_(n−1)   p=2 ,  q=4  pq=8
{x+y=2xy=4Sn=xn+ynSn+1=xn+1+yn+1=(x+y)(xn+yn)xy(xn1+yn1)((x+y)(xn+yn)=xn+1+yn+1+xny+xyn=xn+1+yn+1+xy(xn1+yn1))=(x+y)SnxySn1=2Sn4Sn12Sn=Sn+1+4Sn1p=2,q=4pq=8
Commented by Rasheed Soomro last updated on 12/Aug/16
Very Nice!
VeryNice!

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