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Prove-by-mathematical-induction-that-n-4-4n-2-11-is-divisible-by-16-




Question Number 78948 by TawaTawa last updated on 21/Jan/20
Prove by mathematical induction that.     n^4  + 4n^2  + 11   is divisible by 16
Provebymathematicalinductionthat.n4+4n2+11isdivisibleby16
Commented by john santu last updated on 21/Jan/20
let p(n) = n^4 +4n^2 +11  for n=1 ⇒1+4+11=16 ∣16 (true)  suppose n=k such that   p(k)=k^4 +4k^2 +11 = u(mod 16)  we can proof that for   n=k+1 divisible by 16  k^4 +4k^2 +11+(k+1)^4 +4(k+1)+11  next..
letp(n)=n4+4n2+11forn=11+4+11=1616(true)supposen=ksuchthatp(k)=k4+4k2+11=u(mod16)wecanproofthatforn=k+1divisibleby16k4+4k2+11+(k+1)4+4(k+1)+11next..
Commented by TawaTawa last updated on 21/Jan/20
It is for  k + 1 i don′t get sir
Itisfork+1idontgetsir
Commented by mind is power last updated on 21/Jan/20
error in this  n=2k didn′t worck ever
errorinthisn=2kdidntworckever
Commented by TawaTawa last updated on 21/Jan/20
That means the question is wrong sir?
Thatmeansthequestioniswrongsir?
Commented by mind is power last updated on 21/Jan/20
yeah
yeah
Commented by john santu last updated on 21/Jan/20
and    ...in fact not proved miss  the equation error. k^4 +4k^2 +11  not divisible by 16.  i.q let k=2 ⇒16+16+11≠∣16
andinfactnotprovedmisstheequationerror.k4+4k2+11notdivisibleby16.i.qletk=216+16+11≠∣16
Commented by TawaTawa last updated on 21/Jan/20
God bless you sir
Godblessyousir
Answered by mind is power last updated on 21/Jan/20
first Methode  n=2k  ⇒n^4 +4n^2 +11=16k^4 +16k^2 +11≡11(16)  error sir
firstMethoden=2kn4+4n2+11=16k4+16k2+1111(16)errorsir
Commented by TawaTawa last updated on 21/Jan/20
Sir what of this    n^4  + 4n^2  + 11  is a multiple of  16  for all odd positive integral
Sirwhatofthisn4+4n2+11isamultipleof16foralloddpositiveintegral
Commented by TawaTawa last updated on 21/Jan/20
Help me prove this
Helpmeprovethis
Commented by mind is power last updated on 21/Jan/20
ifn=(2k+1)  n^4 =(2k+1)^4 =16k^4 +32k^3 +24k^2 +8k+1  n^2 =4k^2 +4k+1  n^4 +4n^2 +1=16k^4 +32k^3 +24k^2 +8k+1+4(4k^2 +4k+1)+11  =16k^4 +32k^3 +40k^2 +24k+16  =16(k^4 +2k^3 )+16+8k(5k+3)=8k(5k+3)mod(16)  k(5k+3)=k(5k+5−2)=5k(k+1)−2k=2m  ⇒8k(5k+3)=8.2m=16m=0(16)  ⇒16∣n^4 +4n^2 +11⇒n=(2k+1)is True
ifn=(2k+1)n4=(2k+1)4=16k4+32k3+24k2+8k+1n2=4k2+4k+1n4+4n2+1=16k4+32k3+24k2+8k+1+4(4k2+4k+1)+11=16k4+32k3+40k2+24k+16=16(k4+2k3)+16+8k(5k+3)=8k(5k+3)mod(16)k(5k+3)=k(5k+52)=5k(k+1)2k=2m8k(5k+3)=8.2m=16m=0(16)16n4+4n2+11n=(2k+1)isTrue
Commented by TawaTawa last updated on 22/Jan/20
God bless you sir
Godblessyousir
Commented by mind is power last updated on 22/Jan/20
y′re Welcom Sir Withe pleasur
yreWelcomSirWithepleasur

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