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Find-the-greatest-coefficient-in-the-expansion-of-6-4x-3-




Question Number 62945 by Tawa1 last updated on 27/Jun/19
Find the greatest coefficient in the expansion of   (6 − 4x)^(−3)
Findthegreatestcoefficientintheexpansionof(64x)3
Answered by mr W last updated on 27/Jun/19
(6−4x)^(−3)   =6^(−3) (1−((2x)/3))^(−3)   =6^(−3) Σ_(n=0) ^∞ C_2 ^(n+2) ((2/3))^n x^n   =6^(−3) Σ_(n=0) ^∞ a_n x^n   a_n =C_2 ^(n+2) ((2/3))^n =(((n+2)(n+1))/2)((2/3))^n   a(x)=(((x+2)(x+1))/2)((2/3))^x   a(x) has maximum at x=3.48  ⇒a_n  has maximum at n=3 or 4  a_3 =((5×4)/2)×((2/3))^3 =((80)/(27))  a_4 =((6×5)/2)×((2/3))^4 =((80)/(27))  ⇒x^3  and x^4  term has the largest coef.  which is 6^(−3) ×((80)/(27))=((10)/(729)).
(64x)3=63(12x3)3=63n=0C2n+2(23)nxn=63n=0anxnan=C2n+2(23)n=(n+2)(n+1)2(23)na(x)=(x+2)(x+1)2(23)xa(x)hasmaximumatx=3.48anhasmaximumatn=3or4a3=5×42×(23)3=8027a4=6×52×(23)4=8027x3andx4termhasthelargestcoef.whichis63×8027=10729.
Commented by Tawa1 last updated on 27/Jun/19
God bless you sir
Godblessyousir
Commented by Tawa1 last updated on 29/Jun/19
Sir, how did you get.     3.48   and   3 or 4
Sir,howdidyouget.3.48and3or4
Commented by mr W last updated on 29/Jun/19
look at a(x)=(((x+2)(x+1))/2)((2/3))^x   (x+2)(x+1) is increasing with x,  ((2/3))^x  is decreasing with x,  so an(x) has a maximum at some  point, at this point ((da(x))/dx)=0.  ((da(x))/dx)=(((2x+3))/2)((2/3))^x +(((x^2 +3x+1))/2)((2/3))^x ln ((2/3))=0  (2x+3)+(x^2 +3x+1)ln ((2/3))=0  ln ((2/3))x^2 +[3ln ((2/3))−2]x+[ln ((2/3))−3]=0  ⇒x=((−3ln ((2/3))+2+(√([3ln ((2/3))−2]^2 −4 ln ((2/3)) [ln ((2/3))−3])))/(2ln ((2/3))))=3.48  i.e. a(x) has maximum at x=3.48  a_n  will have maximum when n is  so close to 3.48 as possible, that is  n=3 or n=4.
lookata(x)=(x+2)(x+1)2(23)x(x+2)(x+1)isincreasingwithx,(23)xisdecreasingwithx,soan(x)hasamaximumatsomepoint,atthispointda(x)dx=0.da(x)dx=(2x+3)2(23)x+(x2+3x+1)2(23)xln(23)=0(2x+3)+(x2+3x+1)ln(23)=0ln(23)x2+[3ln(23)2]x+[ln(23)3]=0x=3ln(23)+2+[3ln(23)2]24ln(23)[ln(23)3]2ln(23)=3.48i.e.a(x)hasmaximumatx=3.48anwillhavemaximumwhennissocloseto3.48aspossible,thatisn=3orn=4.
Commented by Tawa1 last updated on 29/Jun/19
God bless you sir
Godblessyousir
Commented by mr W last updated on 13/Jul/19
an other way to find the largest coef.:  a_n =(((n+2)(n+1))/2)((2/3))^n   a_(n−1) =(((n+1)n)/2)((2/3))^(n−1)   a_(n+1) =(((n+3)(n+2))/2)((2/3))^(n+1)   such that a_n  is maximum,   (a_n /a_(n−1) )≥1:  (((n+2)(n+1))/((n+1)n))((2/3))≥1  ((n+2)/n)≥(3/2)  2n+4≥3n  ⇒n≤4  (a_n /a_(n+1) )≥1:  (((n+2)(n+1))/((n+3)(n+2)))((2/3))^(−1) ≥1  ((n+1)/(n+3))≥(2/3)  3n+3≥2n+6  ⇒n≥3  ⇒n=3 or 4
anotherwaytofindthelargestcoef.:an=(n+2)(n+1)2(23)nan1=(n+1)n2(23)n1an+1=(n+3)(n+2)2(23)n+1suchthatanismaximum,anan11:(n+2)(n+1)(n+1)n(23)1n+2n322n+43nn4anan+11:(n+2)(n+1)(n+3)(n+2)(23)11n+1n+3233n+32n+6n3n=3or4

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