Question Number 125697 by aurpeyz last updated on 13/Dec/20
$${what}\:{is}\:{the}\:{largest}\:{coefficient}\:{of}\: \\ $$$$\left(\mathrm{4}+\mathrm{3}{x}\right)^{−\mathrm{5}} ? \\ $$
Answered by mr W last updated on 13/Dec/20
![(4+3x)^(−5) =4^(−5) [1−(−((3x)/4))]^(−5) =(1/(1024))Σ_(k=0) ^∞ C_4 ^(k+4) (−((3x)/4))^k =(1/(1024))Σ_(k=0) ^∞ (−1)^k C_4 ^(k+4) ((3/4))^k x^k =Σ_(k=0) ^∞ a_k x^k a_k =(((−1)^k )/(1024))C_4 ^(k+4) ((3/4))^k positive coef. when k=even=2n negative coef. when k=odd=2n+1 to find the largest coef. we only need to look at positive coefficients, i.e. k=2n. a_(2n) =(1/(1024))C_4 ^(2n+4) ((3/4))^(2n) a_(2(n+1)) =(1/(1024))C_4 ^(2n+6) ((3/4))^(2(n+1)) to find the largest coef. we need to find the smallest n satisfying a_(2n) >a_(2(n+1)) i.e. (1/(1024))C_4 ^(2n+4) ((3/4))^(2n) >(1/(1024))C_4 ^(2n+6) ((3/4))^(2(n+1)) C_4 ^(2n+4) >C_4 ^(2n+6) ((3/4))^2 (((2n+4)!)/(4!(2n)!))>(((2n+6)!)/(4!(2n+2)!))((3/4))^2 ((16)/9)>(((n+3)(2n+5))/((n+1)(2n+1))) 16(2n^2 +3n+1)>9(2n^2 +11n+15) 14n^2 −51n−119>0 n>((51+(√(51^2 +4×14×119)))/(28))≈5.3 the smallest n is 6, i.e. the largest coefficient is a_(12) : a_(12) =(1/(1024))C_4 ^(16) ((3/4))^(12) =((241 805 655)/(4 294 967 296)) you can check this with WolframAlpha:](https://www.tinkutara.com/question/Q125701.png)
$$\left(\mathrm{4}+\mathrm{3}{x}\right)^{−\mathrm{5}} \\ $$$$=\mathrm{4}^{−\mathrm{5}} \left[\mathrm{1}−\left(−\frac{\mathrm{3}{x}}{\mathrm{4}}\right)\right]^{−\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1024}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(−\frac{\mathrm{3}{x}}{\mathrm{4}}\right)^{{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1024}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{k}} {x}^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{k}} {x}^{{k}} \\ $$$${a}_{{k}} =\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{1024}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{k}} \\ $$$${positive}\:{coef}.\:{when}\:{k}={even}=\mathrm{2}{n} \\ $$$${negative}\:{coef}.\:{when}\:{k}={odd}=\mathrm{2}{n}+\mathrm{1} \\ $$$${to}\:{find}\:{the}\:{largest}\:{coef}.\:{we}\:{only} \\ $$$${need}\:{to}\:{look}\:{at}\:{positive}\:{coefficients}, \\ $$$${i}.{e}.\:{k}=\mathrm{2}{n}. \\ $$$${a}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1024}}{C}_{\mathrm{4}} ^{\mathrm{2}{n}+\mathrm{4}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}{n}} \\ $$$${a}_{\mathrm{2}\left({n}+\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{1024}}{C}_{\mathrm{4}} ^{\mathrm{2}{n}+\mathrm{6}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$${to}\:{find}\:{the}\:{largest}\:{coef}.\:{we}\:{need}\:{to} \\ $$$${find}\:{the}\:{smallest}\:{n}\:{satisfying} \\ $$$${a}_{\mathrm{2}{n}} >{a}_{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$${i}.{e}. \\ $$$$\frac{\mathrm{1}}{\mathrm{1024}}{C}_{\mathrm{4}} ^{\mathrm{2}{n}+\mathrm{4}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}{n}} >\frac{\mathrm{1}}{\mathrm{1024}}{C}_{\mathrm{4}} ^{\mathrm{2}{n}+\mathrm{6}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{2}{n}+\mathrm{4}} >{C}_{\mathrm{4}} ^{\mathrm{2}{n}+\mathrm{6}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\frac{\left(\mathrm{2}{n}+\mathrm{4}\right)!}{\mathrm{4}!\left(\mathrm{2}{n}\right)!}>\frac{\left(\mathrm{2}{n}+\mathrm{6}\right)!}{\mathrm{4}!\left(\mathrm{2}{n}+\mathrm{2}\right)!}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{16}}{\mathrm{9}}>\frac{\left({n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\mathrm{16}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)>\mathrm{9}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{11}{n}+\mathrm{15}\right) \\ $$$$\mathrm{14}{n}^{\mathrm{2}} −\mathrm{51}{n}−\mathrm{119}>\mathrm{0} \\ $$$${n}>\frac{\mathrm{51}+\sqrt{\mathrm{51}^{\mathrm{2}} +\mathrm{4}×\mathrm{14}×\mathrm{119}}}{\mathrm{28}}\approx\mathrm{5}.\mathrm{3} \\ $$$${the}\:{smallest}\:{n}\:{is}\:\mathrm{6},\:{i}.{e}.\:{the}\:{largest} \\ $$$${coefficient}\:{is}\:{a}_{\mathrm{12}} : \\ $$$${a}_{\mathrm{12}} =\frac{\mathrm{1}}{\mathrm{1024}}{C}_{\mathrm{4}} ^{\mathrm{16}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{12}} =\frac{\mathrm{241}\:\mathrm{805}\:\mathrm{655}}{\mathrm{4}\:\mathrm{294}\:\mathrm{967}\:\mathrm{296}} \\ $$$$ \\ $$$${you}\:{can}\:{check}\:{this}\:{with}\:{WolframAlpha}: \\ $$
Commented by mr W last updated on 13/Dec/20
Commented by aurpeyz last updated on 13/Dec/20
$${wow}.\:{this}\:{is}\:{so}\:{explicit}.\:{i}\:{cant}\: \\ $$$${appreciate}\:{you}\:{enough}.\:{thanks} \\ $$