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Question Number 122957 by I want to learn more last updated on 21/Nov/20

Commented by mr W last updated on 21/Nov/20

see Q74970

seeQ74970

Answered by mr W last updated on 21/Nov/20

p_k =a^k +b^k +c^k e_1 =a+b+c e_2 =ab+bc+ca e_3 =abc e_(i≥4) =0 e_1 =p_1 =4 2e_2 =e_1 p_1 −p_2 =16−10=6 ⇒e_2 =3 3e_3 =e_2 p_1 −e_1 p_2 +p_3 =12−40+22=−6 ⇒e_3 =−2 0=e_3 p_1 −e_2 p_2 +e_1 p_3 −p_4 ⇒p_4 =−2×4−3×10+4×22=50 0=−e_3 p_2 +e_2 p_3 −e_1 p_4 +p_5 ⇒p_5 =−2×10−3×22+4×50=114 ...... p_n =4p_(n−1) −3p_(n−2) −2p_(n−3) z^3 −4z^2 +3z+2=0 (z−2)(z^2 −2z−1)=0 z=2, 1±(√2) ⇒p_n =a^n +b^n +c^n =2^n +(1+(√2))^n +(1−(√2))^n p_(100) =2^(100) +(1+(√2))^(100) +(1−(√2))^(100) =2^(100) +Σ_(k=0) ^(100) C_k ^(100) [((√2))^k +(−(√2))^k ] =(2^(10) )^(10) +Σ_(k=0) ^(50) C_(2k) ^(100) 2^(k+1) ......

pk=ak+bk+cke1=a+b+ce2=ab+bc+cae3=abcei⩾4=0e1=p1=42e2=e1p1−p2=16−10=6⇒e2=33e3=e2p1−e1p2+p3=12−40+22=−6⇒e3=−20=e3p1−e2p2+e1p3−p4⇒p4=−2×4−3×10+4×22=500=−e3p2+e2p3−e1p4+p5⇒p5=−2×10−3×22+4×50=114......pn=4pn−1−3pn−2−2pn−3z3−4z2+3z+2=0(z−2)(z2−2z−1)=0z=2,1±2⇒pn=an+bn+cn=2n+(1+2)n+(1−2)np100=2100+(1+2)100+(1−2)100=2100+∑100k=0Ck100[(2)k+(−2)k]=(210)10+∑50k=0C2k1002k+1......

Commented by I want to learn more last updated on 21/Nov/20

Wow, thanks sir. Hope for the final answer sir. I appreciate your time sir.

Wow,thankssir.Hopeforthefinalanswersir.Iappreciateyourtimesir.

Commented by Tawa11 last updated on 23/Jul/21

Great

Great

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