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Question Number 21154 by youssoufab last updated on 14/Sep/17
prove: ∀n∈N^∗ ,∀(a,b)∈C^2 , a^(2n+1) +b^(2n+1) = Σ_(k=o) ^(2n) (−1)^k a^k b^(2n−k)
prove:∀n∈N∗,∀(a,b)∈C2,a2n+1+b2n+1=∑2nk=o(−1)kakb2n−k
Commented by alex041103 last updated on 15/Sep/17
What does C^2 mean?? Sorry for the question.
WhatdoesC2mean??Sorryforthequestion.
Commented by sma3l2996 last updated on 15/Sep/17
that′s mean a∈C and b∈C
that′smeana∈Candb∈C
Answered by alex041103 last updated on 15/Sep/17
Let′s start with the expresion (a+b)(Σ_(k=0) ^(2n) (−1)^k a^k b^(2n−k) )= =Σ_(k=0) ^(2n) (−1)^k a^(k+1) b^(2n−k) +Σ_(k=0) ^(2n) (−1)^k a^k b^(2n+1−k) = =(−1)^(2n) a^(2n+1) b^0 +Σ_(k=0) ^(2n−1) (−1)^k a^(k+1) b^(2n−k) +(−1)^0 a^0 b^(2n+1) +Σ_(k=1) ^(2n) (−1)^k a^k b^(2n+1−k) = =a^(2n+1) +b^(2n+1) +Σ_(k=1) ^(2n) (−1)^(k−1) a^k b^(2n+1−k) +Σ_(k=1) ^(2n) (−1)^k a^k b^(2n+1−k) Because (−1)^(k−1) =(((−1)^k )/(−1))=−(−1)^k ⇒(a+b)(Σ_(k=0) ^(2n) (−1)^k a^k b^(2n−k) )= =a^(2n+1) +b^(2n+1) −Σ_(k=1) ^(2n) (−1)^k a^k b^(2n+1−k) +Σ_(k=1) ^(2n) (−1)^k a^k b^(2n+1−k) = =a^(2n+1) +b^(2n+1) Note: The statement a^(2n+1) +b^(2n+1) =Σ_(k=0) ^(2n) (−1)^k a^k b^(2n−k) is False. The true statement is that a^(2n+1) +b^(2n+1) =(a+b)[Σ_(k=0) ^(2n) (−1)^k a^k b^(2n−k) ]
Let′sstartwiththeexpresion(a+b)(∑2nk=0(−1)kakb2n−k)==∑2nk=0(−1)kak+1b2n−k+∑2nk=0(−1)kakb2n+1−k==(−1)2na2n+1b0+∑2n−1k=0(−1)kak+1b2n−k+(−1)0a0b2n+1+∑2nk=1(−1)kakb2n+1−k==a2n+1+b2n+1+∑2nk=1(−1)k−1akb2n+1−k+∑2nk=1(−1)kakb2n+1−kBecause(−1)k−1=(−1)k−1=−(−1)k⇒(a+b)(∑2nk=0(−1)kakb2n−k)==a2n+1+b2n+1−∑2nk=1(−1)kakb2n+1−k+∑2nk=1(−1)kakb2n+1−k==a2n+1+b2n+1Note:Thestatementa2n+1+b2n+1=∑2nk=0(−1)kakb2n−kisFalse.Thetruestatementisthata2n+1+b2n+1=(a+b)[∑2nk=0(−1)kakb2n−k]
Commented by alex041103 last updated on 15/Sep/17
Note: The false statement a^(2n+1) +b^(2n+1) =Σ_(k=0) ^(2n) (−1)^k a^k b^(2n−k) is in fact allways true when a+b=1 or { ((Re(a)+Re(b)=1)),((Im(a)=−Im(b))) :} And also when a+b≠0 and a^(2n+1) +b^(2n+1) =0 (both of those are true for n≠0)
Note:Thefalsestatementa2n+1+b2n+1=∑2nk=0(−1)kakb2n−kisinfactallwaystruewhena+b=1or{Re(a)+Re(b)=1Im(a)=−Im(b)Andalsowhena+b≠0anda2n+1+b2n+1=0(bothofthosearetrueforn≠0)
Commented by youssoufab last updated on 15/Sep/17
am sorry it is: a^(2n+1) +b^(2n+1) =(a+b)Σ_(k=0) ^(2n) (−1)^k a^k b^(2n−k) thanks flr help !
amsorryitis:a2n+1+b2n+1=(a+b)∑2nk=0(−1)kakb2n−kthanksflrhelp!

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