Question Number 114739 by mr W last updated on 20/Sep/20

find the largest and smallest coefficient in (4+3x)^(−5) .

Answered by mr W last updated on 20/Sep/20

(4+3x)^(−5) =4^(−5) (1+((3x)/4))^(−5) =4^(−5) Σ_(k=0) ^∞ (−1)^k C_4 ^(k+4) ((3/4))^k x^k we see the even terms have positive coefficients and the odd terms have always negative coefficients. even terms: k=2n with n=0,1,2,3,... a_(2n) =4^(−5) C_4 ^(2n+4) ((3/4))^(2n) let a_(2n) >a_(2(n+1)) 4^(−5) C_4 ^(2n+4) ((3/4))^(2n) >4^(−5) C_4 ^(2(n+1)+4) ((3/4))^(2(n+1)) (((2n+4)!)/(4!(2n)!))>(((2n+6)!)/(4!(2n+2)!))((3/4))^2 ((16)/9)>(((2n+5)(n+3))/((2n+1)(n+1))) 15n^2 −51n−119>0 n>((51+(√(51^2 +4×15×119)))/(30))≈4.99 ⇒n≥5, i.e. the largest coefficient is for the term x^(10) : a_(10) =4^(−5) C_4 ^(14) ((3/4))^(10) =((1001×3^(10) )/4^(15) ) =((59 108 049)/(1 073 741 824)) odd terms: k=2n+1 a_(2n+1) =−4^(−5) C_4 ^(2n+5) ((3/4))^(2n+1) let a_(2n+1) <a_(2(n+1)+1) −4^(−5) C_4 ^(2n+5) ((3/4))^(2n+1) <−4^(−5) C_4 ^(2n+7) ((3/4))^(2n+3) C_4 ^(2n+5) >C_4 ^(2n+7) ((3/4))^2 ((16)/9)>(((n+3)(2n+7))/((n+1)(2n+3))) 14n^2 −37n−141>0 n>((37+(√(37^2 +4×14×141)))/(28))≈4.8 ⇒n≥5, i.e. the smallest coefficient is for the term x^(11) . a_(11) =−4^(−5) C_4 ^(15) ((3/4))^(11) =−((1365×3^(11) )/4^(16) ) =−((241 805 655)/(4 294 967 396))